make a the subject of relation x=√1/a square-c/d​

Answer:

268m²

Step-by-step explanation:

given,

l= 10m

b= 8m

h= 3m

A = ?

we know,

A = 2 (lb + bh + lh)

= 2 (10m×8m + 8m×3m + 10m×3m)

= 2 ( 80m² + 24m² + 30m² )

= 2 ( 134m² )

= 268m²

Answer:

Hello!

_____________________

The total amount of income tax to withhold is: 35.70 + 34.67 = 70.37$

Step-by-step explanation: Subtract this amount from the salary: 830 - 77.90 = 752.1 $. This is the amount subject to withholding.

The amount subject to withholding is between 521 and 1613$, which means that the income tax to withhold is: 35.70$ + 15% of excess over $521

Therefore, calculate the excess: 752.1 - 521 = 231.1$

Now, calculate the percentage: 231.1 × 15 ÷ 100 = 34.67$

Hope this helped you!

Answer:

The answer to the best prediction is 115.04

Step-by-step explanation:

We have to:

x = 102

They also tell us that:

y = 5.9 + 1.07 * x

If we replace we have:

y = 5.9 + 1.07 * (102)

y = 115.04

Therefore, the best predicted value of ModifyingAbove and with caret given that the older child has an IQ of 102 is 115.04

Answer:(x-5)(x+9)

Step-by-step explanation:

You want two numbers that can give you -45 in multiplication and two numbers that can add to 4 and that is -5 and 9.

Answer:  (x - 5)(x + 9)

If you have to solve, x=5 or x= -9

Step-by-step explanation: You need two numbers that multiply to be 45.

(could be 3 × 15 or 5 × 9) . The difference between the two factors needs to be 4, the coefficient of the middle term.

9 - 5 =4, so use those.  -45 has a negative sign, so one of the factors must be + and the other -  Since the 4 has the + sign, the larger factor has to be + so the difference will be positive.

So (x -5)(x + 9) are your factors. You can FOIL to be sure

x × x += x² . x × 9 = 9x .  -5 ×  x = -5x . -5 × 9 = -45 .

Combine the x terms: 9x -5x = +4x  

Answer:

a) [tex]P(X=2)=(4C2)(0.375)^2 (1-0.375)^{4-2}=0.330[/tex]  

b) [tex]P(X\geq 2)=1-P(X< 2)=1-[P(X=0)+P(X=1)][/tex]

[tex]P(X=0)=(4C0)(0.375)^0 (1-0.375)^{4-0}=0.153[/tex]  

[tex]P(X=1)=(4C1)(0.375)^1 (1-0.375)^{4-1}=0.366[/tex]  

And replacing we got:

[tex]P(X\geq 2)=1-P(X< 2)=1-[0.153+0.366]=0.481[/tex]

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=4, p=0.375)[/tex]  

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

Part a

[tex]P(X=2)=(4C2)(0.375)^2 (1-0.375)^{4-2}=0.330[/tex]  

Part b

[tex]P(X\geq 2)=1-P(X< 2)=1-[P(X=0)+P(X=1)][/tex]

[tex]P(X=0)=(4C0)(0.375)^0 (1-0.375)^{4-0}=0.153[/tex]  

[tex]P(X=1)=(4C1)(0.375)^1 (1-0.375)^{4-1}=0.366[/tex]  

And replacing we got:

[tex]P(X\geq 2)=1-P(X< 2)=1-[0.153+0.366]=0.481[/tex]

Answer:

The volume of the cone is 94.25 in³.

Step-by-step explanation:

The radius of the base is the distance between the center of the circle and the edge of the base, therefore in this case it is equal to 3 in. The volume of a cone is given by:

[tex]V = \frac{\pi*r^2*h}{3}\\V = \frac{\pi*(3)^2*10}{3}\\V = 94.25 \text{ in}^3[/tex]

The volume of the cone is 94.25 in³.

The point that is a solution to the inequality shown in the graph is:

A. (0,5).

The points that are on the region shaded in blue are solutions to the inequality.

(3,2) and (-3,-6) are on the dashed line, hence they are not solutions. Point (5,0) is to the right of the line, hence it is not a solution, and point (0,5) is a solution, meaning that option A is correct.

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Answer:

f(g(0)) = -1

g(f(0)) = -8

using substitution

Answer:

(2n+1)(n+2)

Step-by-step explanation:

Use basic factor pairs, then figure out the two 2s in the factor pairs add with the one to be five. Then just make the answer above.

Answer:

[tex]y=31x^2+77x+41[/tex]

which agrees with option C in your list of possible answers.

Step-by-step explanation:

Since normally functions are represented on the x-y plane, it is common to replace h(x) with the "y" variable of the vertical axis where its values will be represented (plotted). Then the expression can be also written as follows:

[tex]h(x)=31x^2+77x+41\\y=31x^2+77x+41[/tex]

Answer:

9

Step-by-step explanation:

p^2 -4p +4

Let p = -1

(-1)^1 -4(-1) +4

1 +4+4

9

Answer:

The solution:

X = 16 and Y = -6

Step-by-step explanation:

The equations to be solved are:

x+y = 10 ------- equation 1

x+2y = 4 ----------- equation 2

we can multiply equation 1 by -1 to make the value of x and y negative.

This will give us

-x- y = - 10 ------- equation 3

x+2y = 4 ----------- equation 2

We will now add equations 3 and 2 together so that x will cancel itself out.

this will give us

y = -10 +4 = -6

hence, we have the value of y as -6.

To get the value of x, we can put this value of y into any of the equations above.  (I will use equation 1)

x - 6 = 10

from this, we have that x = 4

Therefore, we have our answer as

X = 16 and Y = -6

Answer:

The answer is given below

Step-by-step explanation:

a)

Let us assume Patricia baked x number of cakes. She put half of the cupcakes (i.e x/2) equally into 6 big boxes.

6 big boxes contained [tex]\frac{x}{2}[/tex] cakes, therefore 1 big box would contain [tex]\frac{x}{2}/6=\frac{x}{12}[/tex] cakes.

Let us assume she put the other half into 14 small boxes, therefore each small box would contain [tex]\frac{x}{2}/14=\frac{x}{28}[/tex] cakes.

There were 45 cupcakes in 3 big boxes and 8 small boxes altogether. That is:[tex]3(\frac{x}{12} )+8(\frac{x}{28})=45\\ 84x+96x=15120\\180x=15120\\x=84[/tex]

Therefore Patricia baked 84 cup cakes

b)

She sold all the small boxes and collected $189, i.e she sold 14 small box for $189. Each small box = $189/14 = $13.5

Answer:

They can be seater in 7,257,600 ways

Step-by-step explanation:

Arrangment formula:

Number of ways that n elements can be arranged, that is, distributed in n places is:

[tex]A_{n} = n![/tex]

In this question:

11 seats(driver and other 10).

Only 2 people can drive.

So

The driver can be 2 people.

The other 10 people are arranged in 10 positions.

[tex]T = 2A_{10} = 2*10! = 7257600[/tex]

They can be seater in 7,257,600 ways

Answer:

40

Step-by-step explanation:

Angles in a circle add up to 360 degrees.

135 + 145 + x + x = 360

280 + 2x = 360

2x = 80

x = 40

Answer:

40

Step-by-step explanation:

The sum of the measures of the central angles of a circle is 360 deg.

145 + 135 + x + x = 360

280 + 2x = 360

2x = 80

x = 40

Answer:

a) Null and alternative hypothesis

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=65-1=64[/tex]

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t<-1.565)=0.0612[/tex]

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Critical value approach

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable X represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), p = 0.52

The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        [tex]=P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135[/tex]

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

[tex]=P(X<3)\\\\1-P(X\geq 3)\\\\=1-[{5\choose 3}(0.52)^{3}(1-0.52)^{5-3}+{5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}]\\\\=1-[0.323960832+0.175478784+0.0380204032]\\\\=0.4625399808\\\\\approx 0.4625[/tex]

Thus, the probability that the flight has empty seats is 0.4625.

Answer:

7 to 15, 7:15, 7/15

Step-by-step explanation:

Ratios can be written as:

a to b

a:b

a/b

We want to find the ratio of exercise days to days in the month. She exercises 14 days out of 30 days in the month. Therefore,

a= 14

b= 30

14 to 30

14:30

14/30

The ratios can be simplified. Both numbers can be evenly divided by 2.

(14/2) to (30/2)

7 to 15

(14/2) : (30/2)

7:15

(14/2) / (30/2)

7/15

Answer:

divide both numbers by 14.. the ans is 1: 2

Step-by-step explanation:

input x , output y

if x= x1 then y=y1 and y1 is the only value then it is a function

if we get multiple values of y then it is not a function

Answer:

b. 1.15

Step-by-step explanation:

The z statistics is given by:

[tex]Z = \frac{X - p}{s}[/tex]

In which X is the found proportion, p is the expected proportion, and s, which is the standard error is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Out of 53 teachers who replied to the survey, 13 claim they are satisfied with their job.

This means that [tex]X = \frac{13}{53} = 0.2453[/tex]

She find that the proportion of satisfied teachers nationally is 18.4%.

This means that [tex]p = 0.184[/tex]

Standard error:

p = 0.184, n = 53.

So

[tex]s = \sqrt{\frac{0.184*0.816}{53}} = 0.0532[/tex]

Z-statistic:

[tex]Z = \frac{X - p}{s}[/tex]

[tex]Z = \frac{0.2453 - 0.184}{0.0532}[/tex]

[tex]Z = 1.15[/tex]

The correct answer is:

b. 1.15

We assume that question b is asking for the distribution of [tex] \\ \overline{x}[/tex], that is, the distribution for the average amount of pollutants.

Answer:

a. The distribution of X is a normal distribution [tex] \\ X \sim N(8.6, 1.3)[/tex].

b. The distribution for the average amount of pollutants is [tex] \\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})[/tex].

c. [tex] \\ P(z>-0.08) = 0.5319[/tex].

d. [tex] \\ P(z>-0.47) = 0.6808[/tex].

e. We do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for [tex] \\ \overline{X}[/tex] is also normal because the sample was taken from a normal distribution.

f. [tex] \\ IQR = 0.2868[/tex] ppm. [tex] \\ Q1 = 8.4566[/tex] ppm and [tex] \\ Q3 = 8.7434[/tex] ppm.

Step-by-step explanation:

First, we have all this information from the question:

a. What is the distribution of X?

The distribution of X is the normal (or Gaussian) distribution. X (uppercase) is the random variable, and follows a normal distribution with [tex] \\ \mu = 8.6[/tex] ppm and [tex] \\ \sigma =1.3[/tex] ppm or [tex] \\ X \sim N(8.6, 1.3)[/tex].

b. What is the distribution of [tex] \\ \overline{x}[/tex]?

The distribution for [tex] \\ \overline{x}[/tex] is [tex] \\ N(\mu, \frac{\sigma}{\sqrt{n}})[/tex], i.e., the distribution for the sampling distribution of the means follows a normal distribution:

[tex] \\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})[/tex].

c. What is the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants?

Notice that the question is asking for the random variable X (and not [tex] \\ \overline{x}[/tex]). Then, we can use a standardized value or z-score so that we can consult the standard normal table.

[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] [1]

x = 8.5 ppm and the question is about [tex] \\ P(x>8.5)[/tex]=?  

Using [1]

[tex] \\ z = \frac{8.5 - 8.6}{1.3}[/tex]

[tex] \\ z = \frac{-0.1}{1.3}[/tex]

[tex] \\ z = -0.07692 \approx -0.08[/tex] (standard normal table has entries for two decimals places for z).

For [tex] \\ z = -0.08[/tex], is [tex] \\ P(z<-0.08) = 0.46812 \approx 0.4681[/tex].

But, we are asked for [tex] \\ P(z>-0.08) \approx P(x>8.5)[/tex].

[tex] \\ P(z<-0.08) + P(z>-0.08) = 1[/tex]

[tex] \\ P(z>-0.08) = 1 - P(z<-0.08)[/tex]

[tex] \\ P(z>-0.08) = 0.5319[/tex]

Thus, "the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants" is [tex] \\ P(z>-0.08) = 0.5319[/tex].

d. For the 38 cities, find the probability that the average amount of pollutants is more than 8.5 ppm.

Or [tex] \\ P(\overline{x} > 8.5)[/tex]ppm?

This random variable follows a standardized random variable normally distributed, i.e. [tex] \\ Z \sim N(0, 1)[/tex]:

[tex] \\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex] [2]

[tex] \\ z = \frac{\overline{8.5} - 8.6}{\frac{1.3}{\sqrt{38}}}[/tex]

[tex] \\ z = \frac{-0.1}{0.21088}[/tex]

[tex] \\ z = \frac{-0.1}{0.21088} \approx -0.47420 \approx -0.47[/tex]

[tex] \\ P(z<-0.47) = 0.31918 \approx 0.3192[/tex]

Again, we are asked for [tex] \\ P(z>-0.47)[/tex], then

[tex] \\ P(z>-0.47) = 1 - P(z<-0.47)[/tex]

[tex] \\ P(z>-0.47) = 1 - 0.3192[/tex]

[tex] \\ P(z>-0.47) = 0.6808[/tex]

Then, the probability that the average amount of pollutants is more than 8.5 ppm for the 38 cities is [tex] \\ P(z>-0.47) = 0.6808[/tex].

e. For part d), is the assumption that the distribution is normal necessary?

For this question, we do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for [tex] \\ \overline{X}[/tex] is also normal because the sample was taken from a normal distribution. Additionally, the sample size is large enough to show a bell-shaped distribution.  

f. Find the IQR for the average of 38 cities.

We must find the first quartile (25th percentile), and the third quartile (75th percentile). For [tex]\\ P(z<0.25)[/tex], [tex] \\ z \approx -0.68[/tex], then, using [2]:

[tex] \\ -0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}[/tex]

[tex] \\ (-0.68 *0.21088) + 8.6 = \overline{X}[/tex]

[tex] \\ \overline{x} =8.4566[/tex]

[tex] \\ Q1 = 8.4566[/tex] ppm.

For Q3

[tex] \\ 0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}[/tex]

[tex] \\ (0.68 *0.21088) + 8.6 = \overline{X}[/tex]

[tex] \\ \overline{x} =8.7434[/tex]

[tex] \\ Q3 = 8.7434[/tex] ppm.

[tex] \\ IQR = Q3-Q1 = 8.7434 - 8.4566 = 0.2868[/tex] ppm

Therefore, the IQR for the average of 38 cities is [tex] \\ IQR = 0.2868[/tex] ppm. [tex] \\ Q1 = 8.4566[/tex] ppm and [tex] \\ Q3 = 8.7434[/tex] ppm.

Answer:

when a straight line cuts two or more parallel lines then the angles forming on the side of transversal line exteriorly opposite to eachother is called exterior alternative angle.

for eg if AB //CD and EF is a transversal line meeting the parallel lines at G abd H then the exterior alternative angle are angle EGB = angle CHF and angle AGE=angle DHF are two pairs of exterior alternative angle .

hope its helpful to uh !!!!!!

Answer:

1 or 9.

Step-by-step explanation:

A coefficient is "a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4xy)".

So, in this case, the coefficient of 9x would be 9.

The coefficient of x^2 would be 1.

Hope this helps!

In the given quadratic expression 9x - 20 + x, 1, 9, and -20 are the coefficients.

In a quadratic expression of the standard form ax² + bx + c, x is the variable and a, b, and c are the numeric coefficients.

In the question, we are asked to identify the coefficients from the given quadratic expression 9x - 20 + x².

First, we try to express the given quadratic expression, 9x - 20 + x², in the standard form of a quadratic expression, ax² + bx + c.

Therefore, 9x - 20 + x² = x² + 9x - 20.

Comparing the expression x² + 9x - 20 with the standard form of a quadratic expression ax² + bx + c, we get a = 1, b = 9, c = -20.

We know that in a quadratic expression ax² + bx + c, x is the variable and a, b, and c are the numeric coefficients.

Thus, we can say that in the given quadratic expression 9x - 20 + x², 1, 9, and -20 are the coefficients.

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Answer:

12345678901234567890

Answer:

[tex]95ft^2[/tex]

Step-by-step explanation:

First, note the surfaces we have. We have four triangles and one square base.  Thus, we can find the surface area of each of them and them add them all up.

First, recall the area of a triangle is [tex]\frac{1}{2} bh[/tex]. We have four of them so:

[tex]4(\frac{1}{2} bh)=2bh[/tex]

The base is 5 while the height is 7. Thus, the total surface area of the four triangles are:

[tex]2(7)(5)=70 ft^2[/tex]

We have one more square base. The area of a square is [tex]b^2[/tex]. The base is 5 so the area is [tex]25ft^2[/tex].

The total surface area is 70+25=95.

Answer:

The 12th percentile of the working lifetime is 7.65 years.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation(which is the square root of the variance) [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 10, \sigma = \sqrt{4} = 2[/tex]

12th percentile:

X when Z has a pvalue of 0.12. So X when Z = -1.175.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.175 = \frac{X - 10}{2}[/tex]

[tex]X - 10 = -1.175*2[/tex]

[tex]X = 7.65[/tex]

The 12th percentile of the working lifetime is 7.65 years.

Answer:

5 in x 5 in

Step-by-step explanation:

The area of the rectangle is given by:

[tex]A=x*y=25\\y=\frac{25}{x}[/tex]

Where x and y are the length and width of the rectangle.

The perimeter is:

[tex]P=2x+2y\\P=2x+2*\frac{25}{x}\\ P=2x+\frac{50}{x}[/tex]

The value of x for which the derivate of the perimeter function is zero is the length that yields the smallest perimeter:

[tex]P=2x+\frac{50}{x} \\\\P'=2-\frac{50}{x^2} =0\\2x^2=50\\x=5\ in[/tex]

The value of y is:

[tex]y=\frac{25}{5}\\y=5\ in[/tex]

Therefore, the dimensions that yield the smallest perimeter are 5 in x 5 in.