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3 years ago

Pat can mow the grass on the football field in 45 minutes. Jordan can mow

Mathematics

1 answer:

Basile [38]3 years ago

6 0

You set up the equation like this:

T/45+T/53=1

You find the least common factor of 53 and 45 and times both sides so you end up with this:

53T+45T= least common factor

Then add up 53T and 45T.

98T= least common factor

Then divide 98 from both sides. That will be your answer.

Sorry I didn't have enough time to find the least common factor but it is a number that they both go into. I would check to make sure I'm correct. I hope this helps!

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3 years ago

Finish the pattern 2,4,8,16,_,_,_,_

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3 years ago

John has a rectangular-shaped field whose length is 62.5 yards and width is 45.3 yards. The area of the field is ________ square

oee [108]

Area = 2831.25 square yards

Perimeter =215.6 yards

EXPLANATION

The area and perimeter of a rectangular field are found using the formula for finding the area and perimeter of a rectangle respectively.

That means, area of the rectangular field is given by the formula,

$A= l\times w$

We just have to substitute
$l=62.5$ and $w= 45.3$ into the given formula and evaluate.

This implies that;

$A= 62.5\times 45.3$

This gives the area of the rectangular-shaped field to be;

$A= 2831.25$ square yards.

Now for the perimeter, we use the formula

$P=2w +2l$

Or

$P=2(w +l)$

Substituting the values for the length and width gives,

$P=2(62.5+45.3)$

$\Rightarrow P=2(107.8)$

$\Rightarrow P=215.6$

Hence the perimeter of the rectangular shaped field is 215.6 yards.

8 0

3 years ago

Some one please help!

nasty-shy [4]

Answer:

The answers are 23, 9,200 and 26, 10,400

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard devi

vladimir1956 [14]

Answer:

(a) p-value = 0.043. Null hypothesis is rejected.

(b) p-value = 0.001. Null hypothesis is rejected.

(d) p-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a t-test for single mean is used.

The significance level for the test is α = 0.05.

The decision rule is:

If the p - value is less than the significance level then the null hypothesis will be rejected. And if the p-value is more than the value of α then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

Hₐ: μ > μ₀

The sample size is, n = 11.

The test statistic value is, t = 1.91 ≈ 1.90.

The degrees of freedom is, (n - 1) = 11 - 1 = 10.

Use a t-table t compute the p-value.

For the test statistic value of 1.90 and degrees of freedom 10 the p-value is:

The p-value is:

P (t₁₀ > 1.91) = 0.043.

The p-value = 0.043 < α = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

Hₐ: μ < μ₀

The sample size is, n = 17.

The test statistic value is, t = -3.45 ≈ 3.50.

The degrees of freedom is, (n - 1) = 17 - 1 = 16.

Use a t-table t compute the p-value.

For the test statistic value of -3.50 and degrees of freedom 16 the p-value is:

The p-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The p-value = 0.001 < α = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

Hₐ: μ ≠ μ₀

The sample size is, n = 7.

The test statistic value is, t = 0.83 ≈ 0.82.

The degrees of freedom is, (n - 1) = 7 - 1 = 6.

Use a t-table t compute the p-value.

For the test statistic value of 0.82 and degrees of freedom 6 the p-value is:

The p-value is:

P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.

The p-value = 0.444 > α = 0.05.

The null hypothesis is not rejected at 5% level of significance.

(d)

The alternate hypothesis is:

Hₐ: μ > μ₀

The sample size is, n = 28.

The test statistic value is, t = 2.13 ≈ 2.12.

The degrees of freedom is, (n - 1) = 28 - 1 = 27.

Use a t-table t compute the p-value.

For the test statistic value of 0.82 and degrees of freedom 6 the p-value is:

The p-value is:

P (t₂₇ > 2.12) = 0.022.

The p-value = 0.444 > α = 0.05.

The null hypothesis is rejected at 5% level of significance.

5 0

3 years ago