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3 years ago

ONLY ANSWER CORRECTLY

Mathematics

1 answer:

Nataly [62]3 years ago

5 0

Answer:

C 0.245

Step-by-step explanation:

have a good day/night sorry if wrong may i please have a bralliest

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Multiply and simplify if possible.<br> sqrt 5x(sqrt x-5 sqrt 5)

RSB [31]

Making the assumption that your problem looks like this,
$\sqrt{5x}(\sqrt{x}-5\sqrt{5})$

we use the distributive property to multiply:
$\sqrt{5x}\times \sqrt{x}-\sqrt{5x}\times5\sqrt{5}
\\
\\ \sqrt{5x^2} - 5\sqrt{25x}
\\
\\ x\sqrt{5}-25\sqrt{x}$

5 0

3 years ago

Find the margin of error for a 90% confidence interval when the standard deviation is LaTeX: \sigma= 50????=50 and LaTeX: n = 25

Murrr4er [49]

Answer:

The margin of error for a 90% confidence interval is 16.4

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 25

Standard deviation = 50

$z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64$

Margin of error =

$z_{critical}\times \dfrac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$1.64\times \dfrac{50}{\sqrt{25}} = 16.4$

Thus, the margin of error for a 90% confidence interval is 16.4

8 0

4 years ago

1. W+ 4 = 16

ExtremeBDS [4]

1. W=16-4=12
2. X=-12-7=-19
3. W=6+15=21
5. y=-2+9=7
6. q=35/7=5

Both questions 4 and 7 aren’t written correctly!!

8 0

3 years ago

Birth weights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz. The propor

OLga [1]

Answer:

The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 110, \sigma = 0.15$

The proportion of infants with birth weights between 125 oz and 140 oz is

This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So

X = 140

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{140 - 110}{15}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 125

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{125 - 110}{15}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.9772 - 0.8413 = 0.1359

The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.

4 0

3 years ago

Marshall's mom buys 2 boxes of granola bars each week. each box contains 8 granola bars. if she continues buying 2 boxes each we

sukhopar [10]

She buys 2 boxes each week for a year (there is 52 weeks in a year) = 2 x 52 = 104 boxes per year
if each box contains 8 granola bars...and she buys 104 boxes, then she buys:
104 * 8 = 832 granola bars

6 0

3 years ago