A right rectangular prism has a length of 7 centimeters, a width of 5 centimeters, and a height of 4 centimeters.

What is the value of x? Enter your answer, as a decimal, in the box.

nlexa [21]

Whats the problem for this?

5 0

3 years ago

Amar keeps picking playing cards out of a standard deck of 52 cards in the hopes that he will

lapo4ka [179]

Answer:

The required probability is, 0.78

Step-by-step explanation:

There are, (52 - 13) = 39 cards which are not spade

5 cards can be chosen from the deck of 52 cards in

$^{52}{C}_{5}$ ways

and out of these 5 choices, none of them will be spade in,

$^{39}{C}_{5}$ ways

So,

P(None of the chosen cards is spade)

= $\frac {^{39}{C}_{5}}{^{52}{C}_{5}}$

$\simeq 0.22$

Hence,

P(out of these 5 choices, at least one card is spade)

$\simeq (1 - 0.22)$

= 0.78

4 0

3 years ago

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

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".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

5 0

3 years ago

I honestly dont know the answer for this

babunello [35]

The picture is blurry

6 0

2 years ago

for every girl taking classes at the martial arts schools there are 3 boys who are taking classes at the school. if there are 23

Yuliya22 [10]

You have to be very careful how you set this up. For every girl (who counts as 1) there are 3 boys (who count as 3). The total number in the proportion is 3 + 1 =4. You are given a total, but that is both boys and girls. That's why you need the 4.

3/4 = x / 236 Cross multiply

4x = 3 * 236

4x = 708 Divide by 4

x = 708/4

x = 177

There are 177 boys at the school and 236 - 177 = 59 girls.

3 0

3 years ago