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cupoosta [38]
2 years ago
10

2 * 4/5 Write the product as the product of a whole number and a unit fraction. Enter whole number in 1st box and fraction in 2n

d box
Mathematics
1 answer:
ivolga24 [154]2 years ago
6 0

Answer:

1 whole and 3/5

Step-by-step explanation:

2 x 4/5 = 8/5 = 1 whole and 3/5

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What is the probability of not rolling a 5 on a standard number cube?
Angelina_Jolie [31]
A standard number cube has 6 sides....

probability of rolling a 5 is 1/6......probability of not rolling a 5 is 5/6
8 0
3 years ago
Read 2 more answers
Christopher spends a total of $235.94. He purchases shoes for $80 as well as several sweaters. If each sweater was $25.99, how m
Xelga [282]

$235.94 - 80 = 155.94

155.94 ÷ 25.99 = 6

He bought 6 sweaters

6 0
3 years ago
Find the midpoint between two points on a number line if one of the points is at -7, and the other point is at 12.
frosja888 [35]

Answer:

D

Step-by-step explanation:

The midpoint is the average of the 2 endpoints, that is

midpoint = \frac{-7+12}{2} = \frac{5}{2} = 2.5 → D

3 0
3 years ago
3 times a number plus 5 is breathers than 12. Graph the possible solutions
Korolek [52]

3x12+5 is greater than 12 .

6 0
3 years ago
Consider the probability that no less than 96 out of 145 people will not get the flu this winter. Assume the probability that a
dsp73

Answer:

0.1324 = 13.24% probability that no less than 96 out of 145 people will not get the flu this winter.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses 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 samples 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)}.

In this problem, we have that:

n = 145, p = 0.61

So

\mu = E(X) = np = 145*0.61 = 88.45

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{145*0.61*0.39} = 5.87

Consider the probability that no less than 96 out of 145 people will not get the flu this winter.

More than 95 people, which is the same as 1 subtracted by the pvalue of Z when X = 95. So

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

Z = \frac{95 - 88.45}{5.87}

Z = 1.115

Z = 1.115 has a pvalue of 0.8676

1 - 0.8676 = 0.1324

0.1324 = 13.24% probability that no less than 96 out of 145 people will not get the flu this winter.

6 0
2 years ago
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