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3 years ago

Help please! I've never been good at percents

Mathematics

1 answer:

Igoryamba3 years ago

5 0

Answer:

the answer is: 64.9 but I'm not sure

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A crab-apple juice blend is mixed in a ratio of cranberry to apple 3 to 5

lord [1]

Answer:

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3 years ago

which of the following best describes 4^2/3? a. sq root of 16, b. cube root of 4, c. sq root of four, d. cube root of 16

son4ous [18]

Answer: D) cube root of 16

================================================

Explanation:

The rule we use is

$x^{m/n} = \sqrt[n]{x^m}$

In this case, x = 4, m = 2 and n = 3.

So,

$x^{m/n} = \sqrt[n]{x^m}\\\\\\4^{2/3} = \sqrt[3]{4^2}\\\\\\4^{2/3} = \sqrt[3]{16}\\\\\\$

Showing that the original expression turns into the cube root of 16.

4 0

3 years ago

Tyler has a lizard named Larry. Tyler wants to build a rectangular pen for Larry to run

QveST [7]

Answer:

You can go for 1x9, 2x8, 3x7, 4x6, 5x5.

5x5 is the best as it builds a square, which maximizes space.

20*2.50 = $50.

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3 years ago

Identify the property that the statement illustrates

oksian1 [2.3K]

Answer:

6 units away from 0.

Step-by-step explanation:

when move to 6 you count the units.

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3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

8 0

3 years ago

Help please! I've never been good at percents ​

Help please! I've never been good at percents