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kherson [118]
3 years ago
9

A square and a rectangle have the same area. If the dimensions of the rectangle are 4 ft by 16 ft, how long is a side of the squ

are?
A 10 ft
B. 8 ft
C 32 ft
D. 6ft
E 20 ft
Mathematics
1 answer:
Maru [420]3 years ago
8 0

9514 1404 393

Answer:

  B. 8 ft

Step-by-step explanation:

The area of the rectangle is ...

  A = LW = (16 ft)(4 ft) = 64 ft²

The area of the square in terms of its side length is ...

  A = s²

For the square to have the area of the rectangle, its side length must be ...

  64 ft² = s²

  s = √(64 ft²) = 8 ft

A side of the square is 8 feet long.

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3 years ago
A worker has four different job offers, each with a contract for six years. Assuming the job descriptions are identical,
Phoenix [80]

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Answer:

  $48,000 to start, with 6% raises

Step-by-step explanation:

The starting salaries are within a few percentage points of each other, so the majority of the difference in earnings will come because of the raises. The offer with the largest raise percentage is likely the best.

The earnings total can be figured as the sum of a geometric series with a first term of "starting salary" and a growth factor of (1+raise percentage). For starting salary 's' and raise percentage 'r', the total earnings in 6 years will be ...

  S = s((1+r)^6 -1)/r

Here are the total earnings, in thousands, for each of the offers:

  a) s = 49, r = .03, S = 316.95

  b) s = 51, r = .01, S = 313.75

  c) s = 48, r = .06, S = 334.82 . . . . best offer

  d) s = 50, r = .02, S = 315.41

The worker can earn the most from a starting salary of $48,000 and 6% increases each year.

7 0
3 years ago
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Olivia had $404 in her bank account. She withdrew $311. About how much money does Olivia have in her account now?
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Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

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

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

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3 years ago
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