You plan to make cakes. Each cake requires 1 1/4 cup of sugar. You have 7 1/3 cups of sugar. How many cakes can you make

Divide the fractions by dividing the numerators.

bonufazy [111]

Answer:

3/5

Step-by-step explanation:

You divide 1 and 3. Keep the denominator since it is the same number

Hope this was helpful, Have a Great Day!!

3 0

2 years ago

Read 2 more answers

At a banquet hall, each table requires 18 plates. The kitchen has 387 plates available. How many full tables can be set?

N76 [4]

21 full tables because if you divide the amount of plates you have by the amount of plates needed for each table you’ll get 21.5 but they asked for full tables so that would be 21

7 0

3 years ago

Consider the probability that exactly 90 out of 148 students will pass their college placement exams. Assume the probability tha

Pepsi [2]

Answer:

0.0491 = 4.91% probability that exactly 90 out of 148 students will pass their college placement exams.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given student will pass their college placement exam is 64%.

This means that $p = 0.64$

Sample of 148 students:

This means that $n = 148$

Mean and standard deviation:

$\mu = E(X) = np = 148(0.64) = 94.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{148*0.64*0.36} = 5.84$

Consider the probability that exactly 90 out of 148 students will pass their college placement exams.

Due to continuity correction, 90 corresponds to values between 90 - 0.5 = 89.5 and 90 + 0.5 = 90.5, which means that this probability is the p-value of Z when X = 90.5 subtracted by the p-value of Z when X = 89.5.

X = 90.5

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

$Z = \frac{90.5 - 94.72}{5.84}$