How many two-digit primes have both their digits non-prime? 6 5 4 3 2

HELP ASAP

boyakko [2]

Answer:

4 ounces

Step-by-step explanation:

46/92 is one half so the recipe out of 100 is divided in half

8/2=4

the recipe also doesn't sound to appetizing

7 0

3 years ago

The bottom of a rectangular swimming pool has a perimeter of 142 meters and an area of 1,050 square meters. What are the dimensi

lubasha [3.4K]

Answer:

Step-by-step explanation:

1,129 not 100% sure but i think u add them both together

3 0

4 years ago

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

choli [55]

Answer:

a) The 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

b) $n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51$

And rounded up we have that n=1831

Step-by-step explanation:

Data given and notation

n=2341 represent the random sample taken

X represent the people that they have watched digitally streamed TV programming on some type of device

$\hat p=0.55$ estimated proportion of people that they have watched digitally streamed TV programming on some type of device

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

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

Part a) Confidence interval

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.55 - 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.523$

$0.55 + 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.577$

And the 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p??

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)

And on this case we have that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

And replacing into equation (b) the values from part a we got:

$n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51$

And rounded up we have that n=1831

8 0

4 years ago

HELP with all questions PLEASE.

max2010maxim [7]

10 days for A by 1,000 m = 2 days 5x2=10 And for B it will be 9.75 kilometers on the thrid day because of .75x3=2.25 then put 7,500 meters to 7.5 kilometers then 7.5+2.25=9.75 kilometers on the third day. For C it would be plan B because it would 20 days to get 20 kilometers for A but plan B is two days ahead of A by taking 7.5x3=22.5=20 days 22.5-.75=21.35 21.35-.75=20.60. Hope this helps because it took me a while.

8 0

4 years ago

Help me please I would really appreciate it

o-na [289]

Step-by-step explanation:

The coordinates of the points are P1(2, 2) and P2(10, 2). The distance <em>d</em><em> </em> between these two points are given by

$d = \sqrt{ {(x\frac{}{2} - x\frac{}{1} ) }^{2} + ({y\frac{}{2} - y\frac{}{1}) }^{2} } \\ = \sqrt{ ({10 - 2})^{2} +( {2 - 2)}^{2}} \\ = 8$

7 0

3 years ago