Round 2826 to the nearest hundred.

Answer:

4. ± 3.012

Step-by-step explanation:

Hello!

Assuming that for both variables X₁ and X₂ n₁= n₂ = 16

You need to test at 1% if the variable is significant, this means, if the slope for X₁ is different from zero (β₁≠0) using the t-statistic and the critical value approach.

The hypotheses are:

H₀: β₁= 0

H₁: β₁≠ 0

α: 0.01

[tex]t= \frac{b_1-\beta_1}{Sb_1} ~t_{n_1-3}[/tex]

The degrees of freedom "n₁-3" are determined by the number of parameters that you estimate for the multiple regression, in this case there are three "β₁" "β₂" and "δ²e"

The rejection region for this test is two-tailed, the critical values are:

±[tex]t_{n-3;1-\alpha /2}= t_{13;0.995}= 3.012[/tex]

I hope this helps!

Answer:

x=9,3

Step-by-step explanation:

x²-12x=-27

x²-12x+(12/2)²=-27+(12/2)²

x²-12x+6²=-27+36

(x-6)²=9

x-6=[tex] \frac{ + }{ - } \sqrt{9} [/tex]

x-6=+3 and x-6=-3

x=9 and 3

Answer:

(A) 1199.168

(B) 1503.372

(C) 159.17728

Step-by-step explanation:

(A) To determine the 30th percentile for the number of chocolate chips in the bag, we find the z-score for the 30th percentile.

Found using a z-table or z-calculator, the z-score for the 30th percentile is -0.524

The formula for finding X (the number of items in a given percentile) is:

X = M + Z(S.D.)

Where M is the mean, Z is the specific z-score of the sought percentile and S.D. is the standard deviation.

So for the 30th percentile,

X = 1261 + (-0.524)(118)

X = 1261 - 61.832 = 1199.168

(B) The number of chocolate chips that make up the middle 98% of chips in the bag is

X = 1261 + (2.054)(118)

X = 1261 + 242.372 = 1503.372

(C) For normal distributions, Interquartile range is Q3 - Q1, that is; 3rd quartile minus 1st quartile.

This is within 1.34896 standard deviations of the mean.

IQR = (1.34896)(118)

IQR = 159.17728

Answer:

z= 56°

hope u understood it...

Answer:

Z=56

Step-by-step explanation:

Because i said so

Answer:

69.5309950592 cm²

Step-by-step explanation:

Area of Square:

Area = [tex]Length * Length[/tex]

Area = 18*18

Area = 324 square cm

Area of circle:

Diameter = 18 cm

Radius = 9 cm

Area = [tex]\pi r^2[/tex]

Area = (3.14)(9)²

Area = (3.14)(81)

Area = 254.469004941 square cm

Area of Shaded area:

=> Area of square - Area of circle

=> 324 - 254.469004941

=> 69.5309950592 cm²

Answer:

the correct answer is 2x

Answer:

D. 2x

Step-by-step explanation:

20x² : 1, 2, 4, 5, 10, 20, x

14x : 1, 2, 7, 14, x

The greatest common factor of the polynomial is 2x.

2x(10x² - 7)

Answer:

d) 01-13 = Penguin, 14-34 = Iguana, 35-56 = Parrot, 57-99 & 00 = Turtle

Step-by-step explanation:

13 − 01 + 1 = 13

34 − 14 + 1 = 21

56 − 35 + 1 = 22

99 − 57 + 1 + 1 = 44

Answer:

3

Step-by-step explanation:

To begin with notice that  

        [tex]\displaymode{ \tan(x) = \frac{30}{10 + 30\cot(y)} }[/tex]

From that equation you get that

10 tan(x)  +  30tan(x) cot(x) = 30

therefore

tan(x) + 3 tan(x) cot(x) = 3

Answer:

P [ x > 59000} = 0,6057

Step-by-step explanation:

We assume Normal Distribution

P [ x > 59000} = (x - μ₀ ) /σ/√n

P [ x > 59000} =  (59000 - 60000)/ 3800

P [ x > 59000} = - 1000/3800/√35

P [ x > 59000} = - 1000*5,916 /3800

P [ x > 59000} =  - 5916/3800

P [ x > 59000} = - 1,55

We look for p value for that z score n z-table and find

P [ x > 59000} = 0,6057

Answer:

A) 95% confidence interval for the population mean PEF for children in biomass households = (3.214, 3.386)

95% confidence interval for the population mean PEF for children in LPG households

= (4.125, 4.375)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The result of the hypothesis test is significant, hence, the true average PEF is lower for children in biomass households than it is for children in LPG households.

C) 95% confidence interval for the population mean FEY for children in biomass households = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.375)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Step-by-step explanation:

A) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the z-distribution. This is because although, there is no information provided for the population standard deviation, the sample sizes are large enough for the sample properties to approximate the population properties.

Finding the critical value from the z-tables,

Significance level for 95% confidence interval

= (100% - 95%)/2 = 2.5% = 0.025

z (0.025) = 1.960 (from the z-tables)

For the children in the biomass households

Sample mean = 3.30

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.20

N = sample size = 755

σₓ = (1.20/√755) = 0.0436724715 = 0.04367

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 3.30 ± (1.960 × 0.04367)

CI = 3.30 ± 0.085598

95% CI = (3.214402, 3.385598)

95% Confidence interval = (3.214, 3.386)

For the children in the LPG households

Sample mean = 4.25

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.75

N = sample size = 750

σₓ = (1.75/√750) = 0.063900965 = 0.063901

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 4.25 ± (1.960 × 0.063901)

CI = 4.25 ± 0.125246

95% CI = (4.12475404, 4.37524596)

95% Confidence interval = (4.125, 4.375)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The null hypothesis usually goes against the claim we are trying to test and would be that the true average PEF for children in biomass households is not lower than that of children in LPG households.

The alternative hypothesis confirms the claim we are testing and is that the true average PEF is lower for children in biomass households than it is for children in LPG households.

Mathematically, if the true average PEF for children in biomass households is μ₁, the true average PEF for children in LPG households is μ₂ and the difference is μ = μ₁ - μ₂

The null hypothesis is

H₀: μ ≥ 0 or μ₁ ≥ μ₂

The alternative hypothesis is

Hₐ: μ < 0 or μ₁ < μ₂

Test statistic for 2 sample mean data is given as

Test statistic = (μ₂ - μ₁)/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = 3.30

n₁ = 755

s₁ = 1.20

μ₂ = 4.25

n₂ = 750

s₂ = 1.75

σ = √[(1.20²/755) + (1.75²/750)] = 0.07740

z = (3.30 - 4.25) ÷ 0.07740 = -12.27

checking the tables for the p-value of this z-statistic

Significance level = 0.01

The hypothesis test uses a one-tailed condition because we're testing in only one direction.

p-value (for z = -12.27, at 0.01 significance level, with a one tailed condition) = < 0.000000001

The interpretation of p-values is that

When the p-value > significance level, we fail to reject the null hypothesis and when the p-value < significance level, we reject the null hypothesis and accept the alternative hypothesis.

Significance level = 0.01

p-value = 0.000000001

0.000000001 < 0.01

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that true average PEF is lower for children in biomass households than it is for children in LPG households.

C) For FEY for biomass households,

Sample mean = 2.3 L/s

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation = 0.5

N = sample size = 755

σₓ = (0.5/√755) = 0.0182

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 2.30 ± (1.960 × 0.0182)

CI = 2.30 ± 0.03567

95% CI = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.375)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Hope this Helps!!!

Answer:

12x/2 or 52/2

Step-by-step explanation:

Ok, perimeter is length+length+width+width. 12x/2 and 52/2 could are probably the answers.

Answer:

Null hypothesis = H₀ = There food preferences among vole species are independent of one another.

Alternate hypothesis = H₁ = There is a relationship between voles and food preference.

Expected meadow vole/apple slices = 29.983051

Expected common vole/apple slices = 28.016949

Expected meadow vole/peanut butter-oatmeal = 31.016949

Expected common vole/peanut butter-oatmeal = 28.983051

Chi-square value = χ² = 2.154239

Degree of freedom = 1

Critical value = 3.841

χ² < Critical value

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.

Step-by-step explanation:

He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices.

So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles.

Null hypothesis = H₀ = There food preferences among vole species are independent of one another.

Alternate hypothesis = H₁ = There is a relationship between voles and food preference.

Data collected by Dr. Pagels:

                                              meadow voles     common voles      Row Total

apple slices                                     26                          32                      58

peanut butter-oatmeal                   35                          25                     60

Column Total                                   61                          57                     118

Where 118 is the grand total.

The expected number is given by

Expected = (row total)×(column total)/grand total

Expected meadow vole/apple slices = 58×61/118

Expected meadow vole/apple slices = 29.983051

Expected common vole/apple slices = 58×57/118

Expected common vole/apple slices = 28.016949

Expected meadow vole/peanut butter-oatmeal = 60×61/118

Expected meadow vole/peanut butter-oatmeal = 31.016949

Expected common vole/peanut butter-oatmeal = 60×57/118

Expected common vole/peanut butter-oatmeal = 28.983051

The chi-square statistic value is given by

χ² = Σ(Observed - Expected)²/Expected

χ² = (26 - 29.983051)²/29.983051 + (32 - 28.016949)²/28.016949 + (35 - 31.016949)²/31.016949 + (25 - 28.983051)²/28.983051

χ² = 2.154239

The degrees of freedom is given by

DoF = (row - 1)×(col - 1)

For the given case, we have 2 rows and 2 columns

DoF = (2 - 1)×(2 - 1)

DoF = 1

The given level of significance = 0.05

The critical value from the chi-square table at α = 0.05 and DoF = 1 is found to be

Critical value = 3.841

Conclusion:

Reject H₀ If χ² > Critical value

We reject the Null hypothesis If the calculated chi-square value is more than the critical value.

For the given case,

χ² < Critical value

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.

Answer:

2.26 repeating

Step-by-step explanation:

Convert the fraction to a decimal by dividing the numerator by the denominator.

hope this helpes

be sure to give brainliest

Using the Pythagorean theorem

Hypotenuse = sqrt( 11^2 + 7^2)

= sqrt( 121 + 39)

= sqrt( 160)

= 12.694

Answer:

Step-by-step explanation:

There are 9 possible outcomes for (Bill, Ben)'s cards:

  (4, 4) total 8; (4, 5) total 9; (4, 6) total 10;

  (5, 4) total 9; (5, 5) total 10; (5, 6) total 11;

  (6, 4) total 10; (6, 5) total 11; (6, 6) total 12.

Of these 9 outcomes, 4 have an odd total; 6 are less than 11.

 P(odd) = 4/9

  P(sum < 11) = 2/3

Answer:

Step-by-step explanation:

y + 8 = 0(x - 0)

y + 8 = 0

y = -8

Answer:

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

2¹³

13 x 13

Step-by-step explanation:

We simply find numbers that can multiply to 26 and write out the multiplication to get our answer.

Answer:

  5

Step-by-step explanation:

Substitute the given values and do the arithmetic.

  [tex]0.3y+\dfrac{y}{z}=0.3\cdot 10+\dfrac{10}{5}=3+2=\boxed{5}[/tex]

Answer:

(a)[tex]D(x)=-2,500x+60,000[/tex]

(b)[tex]R(x)=60,000x-2500x^2[/tex]

(c) x=12

(d)Optimal ticket price: $12

Maximum Revenue:$360,000

Step-by-step explanation:

The stadium holds up to 50,000 spectators.

When ticket prices were set at $12, the average attendance was 30,000.

When the ticket prices were on sale for $10, the average attendance was 35,000.

(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)

Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).

[tex]\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500[/tex]

Therefore, we have:

[tex]y=-2500x+b[/tex]

At point (12,30000)

[tex]30000=-2500(12)+b\\b=30000+30000\\b=60000[/tex]

Therefore:

[tex]D(x)=-2,500x+60,000[/tex]

(b)Revenue

[tex]R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2[/tex]

(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.

[tex]R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12[/tex]

The critical value of R(x) is x=12.

(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]

Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.

[tex]R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0[/tex]

Therefore:

Answer:

2.89

Step-by-step explanation:

just do 1.7×1.7=2.89

Answer:

No

Step-by-step explanation:

If they were opposite, they would be equal.

But they are not, (on one of the spaces in between them is smaller and one is larger.)

Hope I have helped you, since it seems I haven't helped "other people" the last time :/

Answer:

No

Step-by-step explanation:

Answer:

(CX3)+10

Step-by-step explanation:

Answer:

c×3+10= j

Step-by-step explanation:

Answer:

3(7) + 2* |7 - 8| - 12 = 11

Step-by-step explanation:

3(7) + 2* |7 - 8| - 12

21 + 2* |-1| - 12

21 + 2* 1 - 12

21 + 2 - 12

23 - 12 = 11

Hope this helps! :)

Answer:

Yup P is the right one having 62.26%

Step-by-step explanation:

Answer:

Yes

Step-by-step explanation:

We can set it up as

P = 33/53

Q = 20/48

R = 54/90

S = 44/83

This is because we are calculating the percent of yellow birds in the total amt. of birds in a specified park.

Now we calculate =>

P = 33/53 = around 0.62

Q = 20/48 = around 0.416

R = 54/90 = 0.6

S = 44/83 = around 0.53

We find that Park P has the greatest percentage and -->

Thus, Park P is our answer and yes, you are correct.

Answer:

It's written as

[tex]89.7 \times {10}^{9} [/tex]

Or

[tex]8.97 \times {10}^{10} [/tex]

Hope this helps you

Answer:

8.97 * 10 ^10

Step-by-step explanation:

We want one nonzero digit to the left of the decimal

8.97

We moved the decimal 10 places to the left

The exponent is positive 10 since we moved 10 places to the left

8.97 * 10 ^10

Answer: [tex]2\sqrt{1+tant}+C[/tex]

Step-by-step explanation:

To integrate means to find the antiderivative of the function. For this problem, we can use u-substitution.

[tex]\int\limits {\frac{dt}{cos^2t\sqrt{1+tant} } } \[/tex]

Let's first use our identities to rewrite the function. Since [tex]\frac{1}{cosx} =secx[/tex], we can use this identity.

[tex]\int\limits {\frac{sec^2t}{\sqrt{1+tant} } } \,[/tex]

[tex]u=\sqrt{1+tant}[/tex]

[tex]du=\frac{sec^2t}{2\sqrt{1+tant} } dt[/tex]

Now that we have u and du, we can plug them back in.

[tex]\int\limits {2} \, du[/tex]

[tex]\int\limits{2} \, du=2u[/tex]

Since we know u, we can plug that in.

[tex]2\sqrt{1+tant}[/tex]

This may seem like the correct answer, but we forgot to add the constant.

[tex]2\sqrt{1+tant}+C[/tex]

Answer:

Fraction of the figure shaded = [tex]\frac{13}{16}[/tex]

Step-by-step explanation:

Ratio of the areas of the given circles are 1 : 4 : 16

Then the radii of the circles will be in the ratio = [tex]\sqrt{1}:\sqrt{4}:\sqrt{16}[/tex]

                                                                             = 1 : 2 : 4

If the radius of the smallest circle = x units

Then the radius of the middle circle = 2x units

and the radius of the largest circle = 4x units

Area of the smallest circle = πx²

Area of the middle circle = π(2x)² = 4πx²

Area of the largest circle = π(4x)²= 16πx²

Area of the region which is not shaded in the middle circle = πx²(4 - 1)

                                                                                                  = 3πx²

Therefore, area of the shaded region = Area of the largest circle - Area of the region which is not shaded

= 16πx² - 3πx²

= 13πx²

Fraction of the figure which is not shaded = [tex]\frac{\text{Area of the shaded region}}{\text{Area of the largest circle}}[/tex]

= [tex]\frac{13\pi x^{2} }{16\pi x^{2} }[/tex]

= [tex]\frac{13}{16}[/tex]

Answer:

(A)Only f(x) and h(x) have y-intercepts.

(C)The minimum of h(x) is less than the other minimums.

(E)The maximum of g(x) is greater than the other maximums.

Step-by-step explanation:

From the table

f(0)=0 and h(0)=0, therefore, Only f(x) and h(x) have y-intercepts. (Option A)

Therefore, the minimum of h(x) is less than the other minimums. (Option C).

Maximum of f(x)=14

Maximum of g(x)=49

Maximum of h(x)=0

Therefore, the maximum of g(x) is greater than the other maximums. (Option E)

Answer: It's B,C, and E

Step-by-step explanation:

Answer:

A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

Step-by-step explanation:

We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;

68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2,  76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.

Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;

                             P.Q.  =  [tex]\frac{(n-1) \times s^{2} }{\sigma^{2} }[/tex]  ~ [tex]\chi^{2} __n_-_1[/tex]

where, s = sample standard deviation = [tex]\sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} }[/tex] = 5.063

            [tex]\sigma[/tex] = population standard deviation

            n = sample of observations = 22

Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.

So, 90% confidence interval for the population standard deviation, [tex]\sigma[/tex] is ;

P(11.59 < [tex]\chi^{2}__2_1[/tex] < 32.67) = 0.90  {As the critical value of chi at 21 degrees  

                                                  of freedom are 11.59 & 32.67}  

P(11.59 < [tex]\frac{(n-1) \times s^{2} }{\sigma^{2} }[/tex] < 32.67) = 0.90

P( [tex]\frac{ 11.59}{(n-1) \times s^{2}}[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{ 32.67}{(n-1) \times s^{2}}[/tex] ) = 0.90

P( [tex]\frac{(n-1) \times s^{2} }{32.67 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1) \times s^{2} }{11.59 }[/tex] ) = 0.90

90% confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1) \times s^{2} }{32.67 }[/tex] , [tex]\frac{(n-1) \times s^{2} }{11.59 }[/tex] ]

                                     = [ [tex]\frac{21 \times 5.063^{2} }{32.67 }[/tex] , [tex]\frac{21 \times 5.063^{2} }{11.59 }[/tex] ]

                                     = [16.48 , 46.45]

90% confidence interval for [tex]\sigma[/tex] = [[tex]\sqrt{16.48}[/tex] , [tex]\sqrt{46.45}[/tex] ]

                                                 = [4.06 , 6.82]

Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].