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tiny-mole [99]
2 years ago
7

for a scavenger hunt, jim's mom distributed a bag of 725 jelly beans evenly into 29 plastic containers and hid then around the y

ard. if, after the hunt, jim has total of 275 jelly beans, then how many of the plastic containers did he find?
Mathematics
2 answers:
igomit [66]2 years ago
8 0

Answer:

11 bags

Step-by-step explanation:

725 jelly beans evenly into 29 plastic containers:

725/29=25  jelly beans in each plastic containers

275/25=11  Jim found eleven bags

ExtremeBDS [4]2 years ago
5 0

Answer:

11 containers

Step-by-step explanation:

His mom evenly divided the jelly beans into containers. There are 725 jelly beans and 29 containers. To find how many jelly beans are in each container, divide 725 by 29.

725/29

25

There are 25 jelly beans in every container.

After the hunt, Jim had 275 jelly beans. There are 25 beans in each container. To find the number of containers, we can divide 275 by 25.

275/25

11

He found 11 containers.

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2 years ago
Suppose that the sitting​ back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 24.4 in.
solong [7]

Answer:

P(X \geq 26.6) = 0.0336, which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

\mu = 24.4, \sigma = 1.2

In this problem, a value x is significantly high if:

P(X \geq x) = 0.01

Using these​ criteria, is a​ back-to-knee length of 26.6 in. significantly​ high?

We have to find the probability of the length being 26.6 in or more, which is 1 subtracted by the pvalue of Z when X = 26.6. So

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

Z = \frac{26.6 - 24.4}{1.2}

Z = 1.83

Z = 1.83 has a pvalue of 0.9664.

1 - 0.9664 = 0.0336

P(X \geq 26.6) = 0.0336, which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

6 0
3 years ago
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