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3 years ago

8

A square garden with a side length 150 m has a square swimming pool in very center with a side length 25 m .Calculate the area o

f the garden.

1 answer:

lana [24]3 years ago

8 0

Total garden area = 150² = 22500m²

Pool area = 25² = 625m²

Actual garden area = 22500 - 625 = 21875m²

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Demetrius received his phone bill in the mail. The bill was for $99.00 and the sales tax was 4.5%. He ended up paying $94.54 for

Margaret [11]

Answer:

false

Step-by-step explanation:

i think the answer is false srry if i am wrong

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2 years ago

Find, correct to four decimal places, the length of the curve of intersection of the cylinder 4x2 1 y2 − 4 and the plane x 1 y 1

charle [14.2K]

<u>Answer-</u> Length of the curve of intersection is 13.5191 sq.units

<u>Solution-</u>

As the equation of the cylinder is in rectangular for, so we have to convert it into parametric form with

x = cos t, y = 2 sin t (∵ 4x² + y² = 4 ⇒ 4cos²t + 4sin²t = 4, then it will satisfy the equation)

Then, substituting these values in the plane equation to get the z parameter,

cos t + 2sin t + z = 2

⇒ z = 2 - cos t - 2sin t

∴ $\frac{dx}{dt} = -\sin t$

$\frac{dy}{dt} = 2 \cos t$

$\frac{dz}{dt} = \sin t-2cos t$

As it is a full revolution around the original cylinder is from 0 to 2π, so we have to integrate from 0 to 2π

∴ Arc length

$= \int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}$

$=\int_{0}^{2\pi}\sqrt{(-\sin t)^{2}+(2\cos t)^{2}+(\sin t-2\cos t)^{2}$

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$

Now evaluating the integral using calculator,

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$ $= 13.5191$

8 0

2 years ago

Raj used 4 3/4 cups of sugar for each batch of muffins. He made 3 1/2 batches. Raj used 16 5/8 cups of sugar altogether.

saul85 [17]

Answer:

It is true!

Step-by-step explanation:

8 0

2 years ago

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A professor finds that the average SAT score among all students attending his college is 1150 ± 150 (μ ± σ). He polls his class

solniwko [45]

Answer:

Step-by-step explanation

Hello!

Be X: SAT scores of students attending college.

The population mean is μ= 1150 and the standard deviation σ= 150

The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.

If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:

H₀: μ = 1150

H₁: μ ≠ 1150

α: 0.05

$Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)$

$Z_{H_0}= \frac{1200-1150}{\frac{150}{\sqrt{25} } } = 1.67$

The p-value for this test is 0.0949

Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.

I hope it helps!

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2 years ago

Cody ate 170.10 grams of chicken for dinner. His sister ate 170.1 grams of chicken. Which statement is true? A. Cody and his sis

8090 [49]

A. They both ate the same because 170.10 is equal to 170.1

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3 years ago

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