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3 years ago

Please helpme if possible withanother thing

Mathematics

2 answers:

Aloiza [94]3 years ago

5 0

Answer:

$\Huge \boxed{256m^3}$

Step-by-step explanation:

To solve this problem, first you have to use the volume formula of $\displaystyle V=L\times W \times H$ .

<h3><u>Given:</u></h3><h3><u /></h3>

$\bigstar\large\boxed{\textnormal{Volume Formula}}\bigstar$

$\displaystyle V=L\times W \times H$

<h3><u>Solution:</u></h3><h3><u /></h3>

Length: 4m

Width: 16m

Height: 4m

Multiply by length times width times height.

$\displaystyle V=L\times W\times H$

$\displaystyle 4\times16\times4=\boxed{256}$

Therefore, the correct answer is (C.) 256m³.

MaRussiya [10]3 years ago

4 0

Volume formula:

V=Width * Height* Length

Width=4

Height=4

Length=16

4*4*16

=16*16

= 256 m^3

Answer: 256 m^3

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4 years ago

There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl

Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

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4 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 437437 green peas and 129129 yellow peas. Use

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Answer:

The null hypothesis: $\mathbf{H_o: p=0.27}$

The alternative hypothesis: $\mathbf{H_1: p \neq 0.27}$

Test statistics : z = −2.30

P-value: = 0.02144

Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27

Step-by-step explanation:

From the given information:

Let's state the null and the alternative hypothesis;

Since The claim is that 27% of the offspring peas will be yellow.

The null hypothesis state that the proportion of offspring peas will be yellow is equal to 0.27.

i.e

$\mathbf{H_o: p=0.27}$

The alternative hypothesis state that the proportion of offspring peas will be yellow is not equal to 0.27

$\mathbf{H_1: p \neq 0.27}$

<u>The test statistics:</u>

we are given 437 green peas and 129 yellow apples;

Hence;

$\hat p = \dfrac{x}{n}$

where ;

$\hat p$ = sample proportion

x = number of success

n = total number of the sample size

$\hat p = \dfrac{129}{437+129}$

$\hat p = \dfrac{129}{566}$

$\mathbf{\hat p = 0.2279}$

Now; the test statistics can be computed as :

$z = \dfrac { \hat p -p }{\sqrt {\dfrac{p(1-p)}{n} } }$

$z = \dfrac {0.2279 -0.27 }{\sqrt {\dfrac{0.27(1-0.27)}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {\dfrac{0.27(0.73)}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {\dfrac{0.1971}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {3.48233216*10^{-4} } }$

$z = \dfrac {-0.043 }{0.01866} }$

z = −2.30

C. P-value

P-value = P(Z < z)

P-value = P(Z< -2.30)

By using the P-value method and the normal distribution as an approximation to the binomial distribution.

from the table of standard normal distribution

move left until the first column is reached. Note the value as –2.0

move upward until the top row is reached. Note the value as 0.30

find the probability value as 0.010724 by the intersection of the row and column values gives the area to the left of

z = -2.30

P- value = 2P(z ≤ -2.30)

P-value = 2 × 0.01072

P - value = 0.02144

Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27

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Answer:

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