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MrRa [10]
3 years ago
5

The length of a rectangle is 7 inches less than twice the width, w, of the rectangle.

Mathematics
1 answer:
Ivan3 years ago
4 0

Answer:

Step-by-step explanation:

Let w represent the width of the rectangle.

The length of a rectangle is 7 inches less than twice the width, w, of the rectangle. It means that the length is (2w - 7) inches

Part A

The function that gives the area as a function of the width is

A(w) = w(2w - 7)

Part B:

If the area of the rectangle is 60 square inches, it means that

w(2w - 7) = 60

2w² - 7w = 60

2w² - 7w - 60 = 0

2w² + 8w - 15w - 60 = 0

2w(w + 4) - 15(w + 4) = 0

2w - 15 = 0 or w + 4 = 0

w = 15/2 or w = - 4

Since w cannot be negative, then

w = 15/2 = 7.5 inches

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Mai and Tyler work on the equation 2/5b+1=-11 together. Mais soulution is b=-25 and Tyler’s is b=-28. Here is their work. Do you
Anika [276]

Answer:

No I don't agree with their solution; both their answers are wrong.

Correct answer is b=-30.

Step-by-step explanation:

Given:

\frac{2}{5}b+1=-11

Now given:

According to Mai b = -25 and According to Tyler b = -28

Now we need to find which of them is correct.

So we will solve the given equation we get;

\frac{2}{5}b+1=-11

Subtracting both side by 1 we get;

\frac{2}{5}b+1-1=-11-1\\\\\frac{2}{5}b =-12

Now Multiplying both side \frac{5}{2} we get;

\frac{2}{5}b\times\frac{5}{2}= -12 \times \frac{5}{2}\\\\b=-6\times 5\\\\b=-30

Hence both of them are incorrect, correct answer is b=-30.

3 0
3 years ago
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

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

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0
3 years ago
The radius of a circle is 7.2 cm. Find the circumference to the nearest tenth.
Lynna [10]

Answer: 45.24

Step-by-step explanation:

6 0
3 years ago
Is the relationship =~ or 180 set up on equation and solver for x find the measure the gíren
larisa86 [58]

Answer:ññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññ

8 0
2 years ago
21 = − 6 x − 3 x + 4 + 4 x
horsena [70]

Answer:

x=-17/5

Step-by-step explanation:

1. Simplify  -6x-3x+4+4x−6x−3x+4+4x  to  -5x+4−5x+4.

21=−5x+4

2. Subtract 44 from both sides.

21−4=−5x

3. Simplify  21-4 to 17.

17=-5x

4. Divided both sides by -5.

-17/5=x

5. Switch sides.

x=-17/5

Hope this helps.

4 0
3 years ago
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