All values equivalent to 1/2

Randomly take 4 cards from a deck of 52 cards, find the probability that the 4 cards are all aces

kompoz [17]

You would divide the number of aces in a deck, which is 4, ny 52 which will give you about 0.0769. As a percentage it will be 7.69%. Round it off to 7.7% if you like. It’s either 7.7% it 7.69%.

3 0

3 years ago

The question is How long did it take each time 18 liters of water were drained?

SVEN [57.7K]

<h2><u><em>It will take 1/4 of a minute to drain 18 liters</em></u></h2>

<h2><u><em>i took a test wit that question</em></u></h2>

3 0

3 years ago

3s - 4 = 8 <br> cant figure it out

Gnesinka [82]

Answer:

3s = 8+4

3s = 12

s = 12/3

s = 4

7 0

3 years ago

What is a linear equation that contains these points?

shepuryov [24]

Answer:

C

Step-by-step explanation:

If you plug in the numbers into that equation you get

1 = -2(2) + 5

1 = -4 + 5

1 = 1

You can also do this again with x = 5 y = -5

-5 = -2(5) + 5

-5 = -10 + 5

-5 = -5

7 0

3 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$