Find the measure of a 3. { > 43 90° {{ 43 = [?] Fnter

Suppose that you have 7 cookies to distribute between 13 children.

Oduvanchick [21]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

3 0

3 years ago

Read 2 more answers

Heeeeeelp please !!!!!!!!

Harlamova29_29 [7]

4440/10 Is the answer

6 0

3 years ago

Read 2 more answers

Pat thinks of a number that is greater than 4 and less than 5.He doubles the number

soldi70 [24.7K]

Answer:

If you know your doubles, then it should be easy to go from 4 + 4 = 8 to 4 + 5 = 9.

Step-by-step explanation:

4 0

3 years ago

So i need help with the question below.

Shkiper50 [21]

Answer:

Option (1)

Step-by-step explanation:

Option (1)

$3\frac{3}{9}+7\frac{6}{11}=10\frac{87}{99}$

$3\frac{3}{9}+7\frac{6}{11}$ $=3+\frac{3}{9}+7+\frac{6}{11}$

$=(3+7)+(\frac{3}{9}+\frac{6}{11})$

$=10+(\frac{3}{9}\times \frac{11}{11})+(\frac{6}{11}\times \frac{9}{9})$

$=10+(\frac{33}{99}+\frac{54}{99})$

$=10+\frac{87}{99}$

$=10\frac{87}{99}$

True.

Option (2)

$2\frac{3}{8}+6\frac{4}{5}=8\frac{12}{40}$

$2\frac{3}{8}+6\frac{4}{5}=2+\frac{3}{8}+6+\frac{4}{5}$

$=(2+6)+(\frac{3}{8}+\frac{4}{5})$

$=(8)+(\frac{3}{8}\times \frac{5}{5} +\frac{4}{5}\times \frac{8}{8})$

$=8+(\frac{15}{40}+\frac{32}{40})$

$=8+\frac{47}{40}$

$=8+\frac{40}{40}+\frac{7}{40}$

$=8+1+\frac{7}{40}$

$=9+\frac{7}{40}$

$=9\frac{7}{40}$

False.

Option(3)

$3\frac{3}{7}+4\frac{2}{3}=7\frac{2}{21}$

$3\frac{3}{7}+4\frac{2}{3}=(3+\frac{3}{7})+(4+\frac{2}{3})$

$=(3+4)+(\frac{3}{7}+\frac{2}{3})$

$=7+(\frac{3}{7}\times \frac{3}{3} +\frac{2}{3}\times \frac{7}{7})$

$=7+(\frac{9}{21}+\frac{14}{21})$

$=7+(\frac{23}{21})$

$=7+(\frac{21}{21}+\frac{2}{21})$

$=8+\frac{1}{21}$

$=8\frac{1}{21}$

False.

Option (4)

$4\frac{5}{6}+5\frac{5}{7}=9\frac{10}{13}$

$4\frac{5}{6}+5\frac{5}{7}=4+\frac{5}{6}+5+\frac{5}{7}$

$=(4+5)+(\frac{5}{6}+\frac{5}{7})$

$=9+(\frac{5}{6}\times \frac{7}{7}+\frac{5}{7}\times \frac{6}{6})$

$=9+(\frac{35}{42}+\frac{30}{42})$

$=9+\frac{65}{42}$

$=9+(\frac{42}{42}+\frac{23}{42})$

$=9+1+\frac{23}{42}$

$=10\frac{23}{42}$

False.

Therefore, Option (1) is the answer.

8 0

3 years ago

The weight of crackers in a box is stated to be 16 ounces. The amount that the packaging machine puts in the boxes is believed t

krek1111 [17]

Answer:

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .