What number should be in the blank in the sequence? 7; 17; 37; 77; ___ ; 317

Answers

Answer 1

Answer:

the answer is 157

Step-by-step explanation:

7 +10= 17

17+20=37

37+40=77

77+80=157

157+160=317

At the beginning you add +10. Every sequence, you need to multiply that number x2. For example: 10 x 2=20...


Related Questions

What is the result of −18⋅16 2/3? Enter the result as an improper fraction and as a mixed number.

Answers

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n

Answers

Answer:

Option C is correct.

The sampling distribution with sample size n=100 will have less variability.

Step-by-step explanation:

Complete Question

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.

A. Both sampling distributions will have the same variability.

B.The sampling distribution with sample size n=20 will have less variability

C. The sampling distribution with sample size n =100 will have less variability

Solution

The central limit theorem allows us to say that as long as

- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.

- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,

1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)

μₓ = μ = 66 inches

2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through

σₓ = (σ/√N)

where σ = population standard deviation = 3.5 inches

N = Sample size

And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.

For sampling distribution with sample size n = 100

σₓ = (3.5/√100) = 0.35 inch

For sampling distribution with sample size n = 20

σₓ = (3.5/√20) = 0.7826 inch

The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.

Hope this Helps!!!

What is the slope of the line with the two
points A(-4, 8) and B(-9, 12)?

Answers

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =

Answers

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

A grocery store manager notices that this month her store sold a total of 597 gallons of ice cream, which represents a decrease of 15% from last month. On the other hand, her store sold 617 pounds of broccoli this month, which represents an increase of 21% from last month. How much ice cream and broccoli did the store sell last month? Round your answers to the nearest integer.

Answers

Answer:

(a)The total sales of ice-cream last month is 702 gallons.

(b)The total sales of broccoli last month is 510 pounds.

Step-by-step explanation:

Part A

Total Sales of gallons of ice cream this month = 597

Since it represents a decrease of 15% of last month's sales

Let the total sales of ice-cream last month =x

Then:

(100-15)% of x =597

85% of x=597

0.85x=597

x=597/0.85

x=702 (to the nearest integer)

The total sales of ice-cream last month is 702 gallons.

Part B

Total Sales of broccoli this month = 617 pounds

Since it represents an increase of 21% of last month's sales

Let the total sales of ice-cream last month =y

Then:

(100+21)% of y =617

121% of y=617

1.21y=617

y=617/1.21

y=510 (to the nearest integer)

The total sales of broccoli last month is 510 pounds.

Suppose that four microchips in a production run of sixty are defective. A sample of six is to be selected to be checked for defects. (a) How many different samples can be chosen

Answers

Answer:

50,063,860 different samples can be chosen

Step-by-step explanation:

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

How many different samples can be chosen

We choose 6 microchips from a set of 60. So

[tex]C_{60,6} = \frac{60!}{6!(60-6)!} = 50063860[/tex]

50,063,860 different samples can be chosen

Which of the following is the solution to 9|x-1|=-45

Answers

Answer:

No solutions.

Step-by-step explanation:

9|x-1|=-45

Divide 9 into both sides.

|x-1| = -45/9

|x-1| = -5

Absolute value cannot be less than 0.

Answer:

No solution

Step-by-step explanation:

=> 9|x-1| = -45

Dividing both sides by 9

=> |x-1| = -5

Since, this is less than zero, hence the equation has no solutions

In the multiplication sentence below, which numbers are the factors? Check
all that apply.
10 x 8 = 80
A. 80
B. 8.
I C. 10

Answers

b and c are the factors because they make the product

Answer:

10 and 8

Step-by-step explanation:

10 and 8 are the factors in this equation because factors are the numbers that are mutiplied together to get the product (The answer to a mutiplication problem) Therefore the factors in this equation are 10 and 8 because those are the numbers that are mutiplied together to get the product.

PLEASE ANSWER FAST !!!
What is the range of the function g for given domain ?

Answers

Answer:

The answer is B

Step-by-step explanation:

Hope this helps.. if not im sorry :(

A research organization keeps track of what citizens think is the most important problem facing the country today. They randomly sampled a number of people in 2003 and again in 2009 using a different random sample of people in 2009 than in 2003 and asked them to choose the most important problem facing the country today from the following​ choices, war,​ economy, health​ care, or other. Which of the following is the correct test to use to determine if the distribution of​ "problem facing this country​today" is different between the two different​ years?

A.

Use a​ chi-square test of homogeneity.

B.

Use a paired​ t-test.

C.

Use a​ two-sample z-test for proportions.

D.

Use a​ chi-square goodness-of-fit test.

Answers

Answer:

Step-by-step explanation:

From the information given, the population is divided into sub groups. Each group would consist of citizens picking a particular choice as the most important problem facing the country. The choices are the different categories. In this case, the null hypothesis would state that the distribution of proportions for all categories is the same in each population. The alternative hypothesis would state that the distributions is different. Therefore, the correct test to use to determine if the distribution of​ "problem facing this country ​today" is different between the two different​ years is

A) Use a​ chi-square test of homogeneity.

Find the following measure for this figure.
Volume =

Answers

Answer:

91 2/3 pi cubic units

Step-by-step explanation:

The formula for the volume of cone is [tex]\dfrac{1}{3}\pi r^2 h[/tex]. Plugging in the given numbers, you get:

[tex]\dfrac{1}{3}\cdot \pi \cdot 5^2 \cdot 11= 91 \ 2/3 \pi[/tex]

Hope this helps!

Answer:

[tex]Volume=\frac{1}{3} \,275\,\pi[/tex] cubic units

Notice that this answer doesn't agree with any of the first three in the list provided via the screenshot

Step-by-step explanation:

Recall the formula for the volume of a cone:

[tex]Volume=\frac{1}{3} Base\,*\,Height[/tex]

In this case the Height is 11 units, and they also give us the radius of the circular base (5 units) from which we can find the circle's base area:

[tex]Area_{circle} = \pi\,R^2\\Area_{circle}=\pi\,(5)^2\\Area_{circle}=25 \pi[/tex]

therefore the total volume becomes:

[tex]Volume=\frac{1}{3} Base\,*\,Height\\Volume=\frac{1}{3} 25\,\pi\,*\,11\\\\Volume=\frac{1}{3} \,275\,\pi[/tex]

HELP ASAP GIVING BRANLIST!!

Answers

Answer:

Question 1: 3 - 5 hours.

Question 2: 0 - 1 hour

Step-by-step explanation:

Question 1: As you can see in the diagram, the guy is moving really slowly and is almost stuck, therefore, it is 3 - 5  hours.

Question 2:  In hours 0 - 1, you can see that the graph is the closest to vertical as it gets.

Can someone please help me??

Answers

Answer : The value of x is 4.1 cm.

Step-by-step explanation :

As we know that the perpendicular dropped from the center divides the chord into two equal parts.

That means,

AB = CB = [tex]\frac{15.6cm}{2}=7.8cm[/tex]

Now we have o calculate the value of x by using Pythagoras theorem.

Using Pythagoras theorem in ΔOBA :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](OA)^2=(OB)^2+(BA)^2[/tex]

Now put all the values in the above expression, we get the value of side OB.

[tex](8.8)^2=(x)^2+(7.8)^2[/tex]

[tex]x=\sqrt{(8.8)^2-(7.8)^2}[/tex]

[tex]x=\sqrt{77.44-60.84}[/tex]

[tex]x=\sqrt{16.6}[/tex]

[tex]x=4.074\approx 4.1[/tex]

Therefore, the value of x is 4.1 cm.

Pet Place sells pet food and supplies including a popular bailed hay for horses. When the stock of this hay drops to 20 bails, a replenishment order is placed. The store manager is concerned that sales are being lost due to stock outs while waiting for a replenishment order. It has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails. The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 bails

Answers

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

Which is the graph |3x-6|=21

Answers

Answer:

it should look like this

what is the length of the line?

Answers

Answer:

root 61

Step-by-step explanation:

You can use the distance formula or draw a triangle with sides 5 and 6

pls helppppp with my math​

Answers

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{2*2}{3*2}\\\\=\frac{5}{6}-\frac{4}{6}\\\\=\frac{5-4}{6}\\\\=\frac{1}{6}[/tex]

Antipsychotic drugs are widely prescribed for conditions such as schizophrenia and bipolar disease. An article reported on body composition and metabolic changes for individuals who had taken various antipsychotic drugs for short periods of time. (a) The sample of 41 individuals who had taken aripiprazole had a mean change in total cholesterol (mg/dL) of 3.55, and the estimated standard error sD n was 3.478. Calculate a confidence interval with confidence level approximately 95% for the true average increase in total cholesterol under these circumstances. (Round your answers to two decimal places.)

Answers

Answer:

95% for the true average increase in total cholesterol under these circumstances

(-2.306 , 9.406)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =41

Mean of the sample(x⁻)  = 3.55

The estimated standard error

                                             [tex]S.E = \frac{S.D}{\sqrt{n} }[/tex]

Given  estimated standard error ( S.E) = 3.478

Level of significance ∝=0.05

Step(ii):-

95% for the true average increase in total cholesterol under these circumstances

[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]

Degrees of freedom

ν= n-1 = 41-1 =40

t₀.₀₅ = 1.6839

95% for the true average increase in total cholesterol under these circumstances

[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]

( 3.55 - 1.6839 ×3.478 ,3.55 + 1.6839 ×3.478 )

(3.55 - 5.856 , 3.55 + 5.856)

(-2.306 , 9.406)

Conclusion:-

95% for the true average increase in total cholesterol under these circumstances

(-2.306 , 9.406)

The production department has installed a new spray machine to paint automobile doors. As is common with most spray guns, unsightly blemishes often appear because of improper mixtures or other problems. A worker counted the number of blemishes on each door. Most doors had no blemishes; a few had one; a very few had two; and so on. The average number was 0.5 per door. The distribution of blemishes followed the Poisson distribution. Out of 10,000 doors painted, about how many would have no blemishes

Answers

Answer:

The numbers of doors that will have no blemishes will be about 6065 doors

Step-by-step explanation:

Let the number of counts by the  worker of each blemishes on the door be (X)

The distribution of blemishes followed the Poisson distribution with parameter  [tex]\lambda =0.5[/tex] / door

The probability mass function on of a poisson distribution Is:

[tex]P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}[/tex]

[tex]P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}[/tex]

The probability that no blemishes occur is :

[tex]P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}[/tex]

[tex]P(X=0) = 0.60653[/tex]

P(X=0) = 0.6065

Assume the number of paints on the door by q = 10000

Hence; the number of doors that have no blemishes  is = qp

=10,000(0.6065)

= 6065

Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.

Answers

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

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segment AB is dilated from the origin to create segment A prime B prime at A' (0, 6) and B' (6, 9). What scale factor was segment AB dilated by?


1/2

2

3

4

Answers

2 is the answer of the question

Answer:

the answer is 3

Step-by-step explanation:

i took the test

Pleassseee hhheeelllppp

Answers

Answer/Step-by-step explanation:

When solving problems like this, remember the following:

1. + × + = +

2. + × - = -

3. - × + = -

4. - × - = +

Let's solve:

a. (-4) + (+10) + (+4) + (-2)

Open the bracket

- 4 + 10 + 4 - 2

= - 4 - 2 + 10 + 4

= - 6 + 14 = 8

b. (+5) + (-8) + (+3) + (-7)

= + 5 - 8 + 3 - 7

= 5 + 3 - 8 - 7

= 8 - 15

= - 7

c. (-19) + (+14) + (+21) + (-23)

= - 19 + 14 + 21 - 23

= - 19 - 23 + 14 + 21

= - 42 + 35

= - 7

d. (+5) - (-10) - (+4)

= + 5 + 10 - 4

= 15 - 4 = 11

e. (-3) - (-3) - (-3)

= - 3 + 3 + 3

= - 3 + 9

= 6

f. (+26) - (-32) - (+15) - (-8)

= 26 + 32 - 15 + 8

= 26 + 32 + 8 - 15

= 66 - 15

= 51

Please answer this question I give brainliest thank you! Number 8

Answers

Answer:

The third options

Step-by-step explanation:

Counting we can see that 10 students went to two or less states, and 10 went to three or more

Answer: half of the students has visited 2 or fewer states.

Explanation: if we count the total amount of students in the survey we can determine that there was a total of 20 students. Half of 20 is 10. Option three is true because if we count the amount of students who have visited 2 or fewer a states it adds up to 10 making that statement true

angle ∠DAC= angle ∠BAD. What is the length of BD? Round to one decimal place.

Answers

Answer: 3.9

Step-by-step explanation: Khan Academy

The length of BD if The angle ∠ DAC is equal to the angle ∠ BAD is 3.92.

What is the triangle?

Three straight lines coming together create a triangle. There are three sides and three corners on every triangle (angles). A triangle's vertex is the intersection of two of its sides. Any one of a triangle's three sides can serve as its base, however typically the bottom side is used.

Given:

The angle ∠ DAC = angle ∠ BAD

As we can see that the triangle BAD and triangle DAC are similar By SAS similarity,

AC / AB = CD / BD  (By the proportional theorem of similarity)

5.6 / 5.1 = 4.3 / BD

1.09 = 4.3 / BD

BD = 4.3 / 1.09

BD = 3.92

Thus, the length of BD is 3.92.

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Find the area of a triangle that has the base of 5 inches and a height of 3 3/4 inches

Answers

Answer:

9.375 in^2

Step-by-step explanation:

Terry has a number cube that is numbered from 1 to 6. She rolls the cube 50 times. Which equation can be used to predict the number of times that she will roll a number that is greater than 4? P (number greater than 4) = StartFraction 1 over 6 EndFraction (50) P (number greater than 4) = StartFraction 2 over 6 EndFraction (50) P (number greater than 4) = StartFraction 3 over 6 EndFraction (50) P (number greater than 4) = StartFraction 4 over 6 EndFraction (50)

Answers

Answer:

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

3z/10 - 4 = -6
someone help?

Answers

Answer:

[tex]z=-\frac{20}{3}[/tex]

Step-by-step explanation:

[tex]\frac{3z}{10}-4=-6\\\\\frac{3z}{10}-4+4=-6+4\\\\\frac{3z}{10}=-2\\\\\frac{10\cdot \:3z}{10}=10\left(-2\right)\\\\3z=-20\\\\\frac{3z}{3}=\frac{-20}{3}\\\\z=-\frac{20}{3}[/tex]

Best Regards!

HELP ASAP!!!The first picture is what each variables equal too

Answers

Answer:

Just replace the variables with the number

d5

c4 (uh oh)

a2

b-3

f-7

d-c = 5 - 4 = 1

1/3 - 4(ab+f)

2 x -3 = -6

-6 + -7 = -13

-13 x 4 = -52

1/3 - -52 = 1/3 + 52 =

52 1/3

Hope this helps

Step-by-step explanation:

ali's typing rate between 8:00 am and noon is 48 words per minute . after lunch a lunch break, Ali's typing rate between 1:00 pm and 4:00 pm is 2,040 words per hour . what is Ali's average typing rate per minute for the whole time she works?

Answers

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

4 hours= 4*60 min= 240 min

She types:

240*48= 11520 words

After lunch she works:

4 hours

She types:

4*2040= 8160 words

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

A 2011 survey, by the Bureau of Labor Statistics, reported that 91% of Americans have paid leave. In January 2012, a random survey of 1000 workers showed that 89% had paid leave. The resulting p-value is .0271; thus, the null hypothesis is rejected. It is concluded that there has been a decrease in the proportion of people, who have paid leave from 2011 to January 2012. What type of error is possible in this situation?

Answers

Answer:

Is possible to make a Type I error, where we reject a true null hypothesis.

Step-by-step explanation:

We have a hypothesis test of a proportion. The claim is that the proportion of paid leave has significantly decrease from 2011 to january 2012. The P-value for this test is 0.0271 and the nunll hypothesis is rejected.

As the conclusion is to reject the null hypothesis, the only error that we may have comitted is rejecting a true null hypothesis.

The null hypothesis would have stated that there is no significant decrease in the proportion of paid leave.

This is a Type I error, where we reject a true null hypothesis.

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