1.5 * 10 exponent -5 covert to decimal form

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

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

Answer:

a. 0.0668

b. 0.9545

Step-by-step explanation:

We have the following information:

mean (m) = 10000

standard deviation (sd) = 100

(a)

We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:

P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]

P (x> 10150) = P (z> 1.5)

P (x> 10150) = 1 - P (z <1.5)

P (x> 10150) = 1 - 0.9332 (attached table)

P (x> 10150) = 0.0668

Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668

(b)

We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:

P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]

P (9800 <x <10200) = P (-2 <z <2)

P (9800 <x <10200) = P (z <2) - P (z <-2)

P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)

P (9800 <x <10200) = 0.9545

the proportion of pieces that would expect to scrap is 0.9545

Answer:

9.375 in^2

Step-by-step explanation:

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

Answer:

a)[tex]A(t)=2000e^{0.085t}[/tex]

b)[tex]A'(t)=170e^{0.085t}[/tex]

c)$3059.1808

d)t=4.77 years

e) The balance growing is $254.99/year

Step-by-step explanation:

We are given that Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously.

Principal = $2000

Rate of interest = 8.5%

a) What is the formula for​ A(t), the balance after t​ years? ​

Formula [tex]A(t)=Pe^{rt}[/tex]

So,[tex]A(t)=2000e^{0.085t}[/tex]

B)What differential equation is satisfied by​ A(t), the balance after t​ years?

So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]

[tex]A'(t)=170e^{0.085t}[/tex]

c)How much money will be in the account after 5 ​years? ​

Substitute t = 5 in the formula "

[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]

d)When will the balance reach ​$3000​?

Substitute A(t)=3000

So, [tex]3000=2000e^{0.085t}[/tex]

t=4.77

The balance reach $3000 in 4.77 years

e)How fast is the balance growing when it reaches ​$3000​?

Substitute the value of t = 4.77 in derivative formula :

[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]

Hence the balance growing is $254.99/year

*see attachment showing the 2 boxes

Answer:

3 ft²

Step-by-step explanation:

==>Given:

Box J with the following dimensions:

L = 4.5ft

W = 3ft

H = 2ft

Box F:

L = 3ft

W = 3ft

H = 3ft

==>Required:

Difference between the surface area of box J and box F

==>Solution:

Surface area = 2(WL + HL + HW)

=>S.A of box J = 2(3*4.5 + 2*4.5 + 2*3)

= 2(13.5 + 9 + 6)

= 2(28.5)

S.A of box J = 57 ft²

=>S.A of box F = 2(3*3 + 3*3 + 3*3)

= 2(9 + 9 + 9)

= 2(27)

S.A of box F = 54 ft²

Difference between box J and box F = 57 - 54 = 3 ft²

The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .

The Laplace transform exist when s > 0 .

Here, the given function is f(t) = sin²(wt) .

The Laplace transform of the the function f(t),

F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }

F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }

F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }

F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]

Next,

The above Laplace transform exist if s > 0 .

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Answer:

[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=40-1=39[/tex]  

Thep value for this case would be given by:

[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.

Step-by-step explanation:

Information provided

[tex]\bar X=5.2[/tex] represent the sample mean

[tex]s=0.8[/tex] represent the sample standard deviation

[tex]n=40[/tex] sample size  

[tex]\mu_o =5[/tex] represent the value to verify

[tex]\alpha=0.05[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value

Hypothesis to test

We want to verify if the true mean is higher than 5, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 5[/tex]  

Alternative hypothesis:[tex]\mu > 5[/tex]  

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

Replacing the info given we got:

[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=40-1=39[/tex]  

Thep value for this case would be given by:

[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.

Answer:

A' - (8,2)

B' - (5,1)

C' - (4,2)

D' - (4,5)

Step-by-step explanation:

See attachment for the missing figure.

We can see that the vertices of the polygon ABCD have coordinates A(4,6), B(5,3), C(4,2) and D(1,2)

Polygon ABCD is rotated 90° clockwise about point C to form polygon A′B′C′D′ (see attached diagram), then

A'(8,2);

B'(5,1);

C' is the same as C, thus, C'(4,2);

D'(4,5).

Answer:

0.03125 = 3.125% probability that the person flipped 5 heads

Step-by-step explanation:

For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [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]

And p is the probability of X happening.

Five coins:

This means that n = 5.

Fair coin:

Equally as likely to be heads or tails, so p = 0.5.

What is the probability that the person flipped 5 heads?

This is P(X = 5).

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

[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]

0.03125 = 3.125% probability that the person flipped 5 heads

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!

Answer:

2/5

Step-by-step explanation:

Answer:

x= 8.00 Interest rate on $14000

y= 7.50 Interest rate on $7000

Step-by-step explanation:

Let interest rate of $14000 be x%

and Interest rate for $7000 be y %

According to the first condition

14000 * x% - 7000 * y% = 595

multiply by 100

14000x-7000y = 59500

/700

20x-10y=85.................(1)

II condition

x%=y%+0.5%

x=y+0.5

x-y=0.5..................................(2)

solve (1) & (2)

20 x -10 y = 85 .............1

Total value

1 x -1 y = 0.50 .............2

Eliminate y

multiply (1)by 1

Multiply (2) by -10

20.00 x -10.00 y = 85.00

-10.00 x + 10.00 y = -5.00

Add the two equations

10.00 x = 80.00

/ 10.00

x = 8.00

plug value of x in (1)

20.00 x -10.00 y = 85.00

160.00 -10.00 y = 85.00

-10.00 y = 85.00 -160.00

-10.00 y = -75.00

y = 7.50

x= 8.00 Interest rate on $14000

y= 7.50 Interest rate on $7000

Answer: B.

x ≈2.5

Step-by-step explanation:

[tex]-\left(u\right)^{-1}-6=-u+10[/tex]

[tex]u=8-\sqrt{65},\:u=8+\sqrt{65}[/tex]

[tex]x=\frac{\ln \left(8+\sqrt{65}\right)}{\ln \left(3\right)}[/tex]

x=2.52...

Answer:

x=2.5

Step-by-step explanation:

Answer:

114

Step-by-step explanation:

Answer:

144Step-by-step explanation:

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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Answer:

r = 7.5

Step-by-step explanation:

Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]

Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!

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

Answer:

a) Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Step-by-step explanation:

a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:

Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Answer:

Option A is correct.

A uniform distribution.

Step-by-step explanation:

Complete Question

T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:

A) Uniform Distribution

B) Continuous Distribution

C) Poisson Distribution

D) Relative Frequency Distribution

Solution

A uniform distribution is one in which all the variables have the same probability of occurring.

It is also known as a rectangular distribution, as every portion of the sample space has an equal chance of occurring, with equal length on the probability curve, leading to a rectangular probability curve.

And for this question, 6 different models of phones sell an equal number, hence, the probability of selling each model is equal to one another, hence, this is evidently a uniform distribution.

Hope this Helps!!!

Answer:

A. [tex](3b+8)^2[/tex]

Step-by-step explanation:

[tex]9b^2+48b +64\\=(3b)^2 + 2\times 3b\times 8 +(8)^2\\=(3b+8)^2[/tex]

Answer:

The 95% confidence interval is 2.5 < u <3.1.

Step-by-step explanation:

The provided sample mean is X = 2.8 and the sample standard deviation is s = 1.2, and the sample size is n = 64.

1. Null and Alternative Hypotheses:

The following null and alternative hypotheses need to be tested:

H0  u = 3

Ha: u < 3

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

2. Rejection Region Based

on the information provided, the significance level is alpha = 0.05, and the critical value for a left-tailed test is t c = -1.669.

The rejection region for this left-tailed test is R = t : t < -1.669

3. Test Statistics

The t-statistic is computed as follows:

t = (X - uo)/[s/n^(1/2)] =

replacing

t = (2.8 - 3)/ [1.2/64 ^(1/2)]

t =-1.333  

4. Decision about the null hypothesis

Since it is observed that t = -1.333 > t c = -1.669, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0936, and since p= 0.0936 => 0.05, it is concluded that the null hypothesis is not rejected.

5. Conclusion It is concluded that the null hypothesis H0 is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 3, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 2.5 < u <3.1.

Answer:

tiStep-by-step explanaon:

Answer:

[0, positive infinity)

Step-by-step explanation:

The domain is all x values a graph inputs. In a square root function, you cannot have negative inputs as it will turn out imaginary numbers. Therefore, your domain is all values of x above and including 0.

Answer: d on Ed

Step-by-step explanation:

Just took the test

Answer:

I am assuming the underlined value is 25%. It is a parameter

Step-by-step explanation:

The value is is a parameter. This is because the parameter is a value that describes the population.

The survey carried out was a national survey of which there were 25% respondents who reported that someone had offered, sold, or given them an illegal drug on school property. It is not a statistics because a sample was not taken out of the population and a survey made on the sample.

The underlined 25% value is the value that summarizes the entire population of high school students

Answer:

The degrees of freedom are given by:

[tex]df=n-1=317-1=316[/tex]

And replaicing we got:

[tex]29-2.1=26.9[/tex]    

[tex]29+2.1=31.1[/tex]    

The 95% confidence interval would be between 26.9 and 31.1

Step-by-step explanation:

Information given

[tex]\bar X= 29[/tex] represent the sample mean

[tex]\mu[/tex] population mean

s represent the sample standard deviation

[tex] ME= 2.1[/tex] represent the margin of error

n represent the sample size  

Solution

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

And this formula is equivalent to:

[tex] \bar X \pm ME[/te]x

The degrees of freedom are given by:

[tex]df=n-1=317-1=316[/tex]

And replaicing we got:

[tex]29-2.1=26.9[/tex]    

[tex]29+2.1=31.1[/tex]    

The 95% confidence interval would be between 26.9 and 31.1

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

Answer:

  $6534.42

Step-by-step explanation:

Put the given values into the simple interest formula and solve for the remaining variable.

  A = P(1 +rt)

where P is the principal invested, r is the annual rate, and t is the number of years.

  $7000 = P(1 +0.095(9/12)) = 1.07125P

  $7000/1.07125 = P ≈ $6534.42

The value that must be invested is $6534.42.

Answer:

domain = {-5, -3, -2, 4, 8, 12}

Step-by-step explanation:

The domain is the set containing the x-coordinates of all ordered pairs.

domain = {-2, -3, 12, 8, 4, -5}

If you'd like, you can put the numbers in ascending order:

domain = {-5, -3, -2, 4, 8, 12}

Answer:

His weekly exercise time goal is 168 minutes.

Step-by-step explanation:

This question can be solved using a rule of three.

42 minutes is 25% = 0.25 of the total

x minutes is 100% = 1 of the total.

Then

42 minutes - 0.25

x minutes - 1

[tex]0.25x = 42[/tex]

[tex]x = \frac{42}{0.25}[/tex]

[tex]x =  168[/tex]

His weekly exercise time goal is 168 minutes.