PLEASE HELP ME PLEASE !

15(3x - 4y)<br> Use Distributive property

MariettaO [177]

45x -4y is the answer

5 0

2 years ago

Read 2 more answers

Hurrryyyyy upppppp plzzzz helpppp meeeeee rightttt nowwwwwwww

bonufazy [111]

Answer:

It’s the third one

Step-by-step explanation:

-1.1x+6.4>-1.3

-1.1x>-1.3-6.4

-1.1x>-7.7 multiplied by -1 gives 1.1x<7.7

Divided by 1.1 you get x<7

7 0

3 years ago

A nurse at a local hospital is interested in estimating the birth weight of infants. how large a sample must she select if she d

enyata [817]

For large sample confidence intervals about the mean you have:

xBar ± z * sx / sqrt(n)

where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size

We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.

z * sx / sqrt(n) = width.

so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348

The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758

Since n must be integer valued we need to take the ceiling of this solution.
n = 22

7 0

3 years ago

Estimate of 0.75 divided by 3.15

Ganezh [65]

0.3333333333333333333

4 0

3 years ago

My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

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

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: