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3 years ago

Madi bought 10 quarts of ice cream how many gallons and quarts of ice cream did Madi buy explain how you found your answer

Mathematics

1 answer:

Afina-wow [57]3 years ago

7 0

The answer is 2 gallons and 2 quarts.

1 gal=4 quarts
Solution

10/4=2 R 2

Final Answer, 2 gallons and 2 quarts.

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Evaluate 12÷3·(-3) + |-8| -20 -4 20/3

olasank [31]

Answer:

-4

Step-by-step explanation:

6 0

3 years ago

A. Show

nydimaria [60]

a. Recall the double angle identities:

$\sin^2x=\dfrac{1-\cos2x}2$

$\cos^2x=\dfrac{1+\cos2x}2$

Then

$\sin^2x\cos^2x=\dfrac{(1-\cos2x)(1+\cos2x)}4=\dfrac{1-\cos^22x}4$

Applying the identity again, we have

$\sin^2x\cos^2x=\dfrac{1-\frac{1+\cos4x}2}4=\dfrac{2-(1+\cos4x)}8=\dfrac{1-\cos4x}8$

as required.

b. Using the result from part (a),

$\sin^2x\cos^2x=\dfrac{1-\cos4x}8=\dfrac{2-\sqrt2}{16}$

$\implies\cos4x=\dfrac1{\sqrt2}$

$\implies4x=\dfrac\pi4+2n\pi\text{ or }4x=-\dfrac\pi4+2n\pi$

(where $n$ is any integer)

$\implies\boxed{x=\pm\dfrac\pi{16}+\dfrac{n\pi}2}$

8 0

3 years ago

4+6x64 please i have to go to school tomorrow

leva [86]

Answer:

388

Step-by-step explanation:

4+6x64

multiply 6 and 64

6 times 64 = 384

Add 4 to 384

384+4=388

6 0

3 years ago

Read 2 more answers

I need help this is needed by tomorrow morning :)

zhuklara [117]

I think it’s 1/48 :)

4 0

3 years ago

Read 2 more answers

A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 80% confidence inter

jasenka [17]

Answer:

The 80% confidence interval for the mean

(4.6199 , 4.7801)

Step-by-step explanation:

Explanation:-

Assuming that the population standard deviation for the number of energy drinks consumed each week is 1

Given the Population standard deviation 'σ' = 1

The study found that for a sample of 256 teenagers the mean number of energy drinks consumed per week is 4.7

given sample size 'n' = 256

mean of the sample 'x⁻' = 4.7

confidence interval for the mean

The 80% confidence interval for the mean is determined by

$(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} + Z_{\alpha }\frac{S.D}{\sqrt{n} } )$

the z-score of 80% level of significance = 1.282

$(4.7 - 1.282\frac{1}{\sqrt{256} } , 4.7 + 1.282\frac{1}{\sqrt{256} } )$

(4.7 - 0.0801 , 4.7 +0.0801)

(4.6199 , 4.7801)

Conclusion:-

The 80% confidence interval for the mean

(4.6199 , 4.7801)

7 0

4 years ago