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Arturiano [62]
3 years ago
7

14/15 divided by 1 1/20=?

Mathematics
2 answers:
aalyn [17]3 years ago
5 0
This is the answer to your question. I hope this helps!

marusya05 [52]3 years ago
5 0
14/15 x 20/11 multiply the reciprocal
14/3 4/11 the greatest common factor is 5 divide 15 & 20 by 5
14/3 x 4/11 multiply both fractions
56/33
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Answer: A.

Step-by-step explanation:

Equation: V=lwh

5 × 10 × 15 = 750 cubic cm

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The height of a street light is 25 feet. It casts a 20−foot shadow. At the same time, a man standing next to the street light ca
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Write a linear function f with f(-3) = 1 and f(13) = 5
GalinKa [24]
D = (5-1)/(13-(-3)) = 0.25

f(13) = 13d + x = 13 * 0.25 + a = 3.25 + a = 5
a = 1.75

f(x) = 1.75 + 0.25x
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4 years ago
Which ordered pair is a solution of the equation y=x-2
lora16 [44]

Answer:

A. (3 , 1)

Step-by-step explanation:

Note the placement of x and y for points:

(x , y)

Plug in the corresponding numbers for the corresponding variables:

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1 = 1 (True).

A. (3 , 1) is your answer.

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6 0
2 years ago
Read 2 more answers
A golfer would like to test the hypothesis that the variance of his golf score equals 10.0. A random sample of 28 rounds of golf
Ratling [72]

Answer:

Step-by-step explanation:

Hello!

Interest hypothesizes is that the variance of the golfer's scores equals to δ²= 10.0.

A random sample of 28 rounds of golf had a sample standard deviation S= 3.5

The statistics hypotheses are:

H₀: δ²= 10.0

H₁: δ²≠ 10.0

α: 0.05

To conduct a hypothesis test for the population variance you have to work using the Chi-Square distribution, this test is two-tailed so you will have two critical values. Between these two values is defined as the "not rejection region" and below and above them lies the "rejection region"

Lower critical value: X^2_{n-1; \alpha /2}= X^2_{27; 0.05}= 16.928

Upper critical value: X^2_{n-1;1-\alpha /2}= X^2_{27;0.95}= 40.113

If X^2_{H_0} ≤ 16.928 or X^2_{H_0} ≥ 40.113, the decision is to reject the null hypothesis.

If 16.928 < X^2_{H_0} < 40.113, the decision is to not reject the null hypothesis.

X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}

X^2_{H_0}= \frac{(28-1)12.25}{10} = 33.075

The calculated X^2_{H_0} is between the two critical values, so the decision is to not reject the null hypothesis. We can conclude that the population variance fro this golfers play score is 10.

I hope this helps!

5 0
3 years ago
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