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Sladkaya [172]
3 years ago
5

3+2+7=7+2+3 TYpe the Values?

Mathematics
2 answers:
Tanzania [10]3 years ago
6 0
That's 12 = 12. What do you want to know about that?
Slav-nsk [51]3 years ago
6 0
Its 12 =12. hope this helped
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Angle 3 and Angle 6 are an example of which type of angle pair?
Elena-2011 [213]

Answer:

Alternate Interior Angles

5 0
3 years ago
five years ago a man was seven times as old as his son. five years hence the gather would be three times as old as his son. find
babymother [125]
Father = x
son = y
x-5 = 7(y-5)
x +5 = 3(y+5) solve the system

x-5 = 7y-35
x+5 = 3y +15

x-5 = 7y-35
-x-5 = -3y -15
-10 = 4y -50
+50 +50
40 = 4y
10 = y

Solve for X
x+5 = 3(10) + 15
x+5 = 45
x = 40

The father is 40yrs old and the son is 10yrs old.
5 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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 16.15, \sigma = 0.3, n = 50, s = \frac{0.3}{\sqrt{50}} = 0.0424

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

This is 1 subtracted by the pvalue of Z when X = 16. So

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

By the Central Limit Theorem

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

Z = \frac{16 - 16.15}{0.0424}

Z = -3.54

Z = -3.54 has a pvalue of 0.0001

1 - 0.0001 = 0.9999

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

3 0
3 years ago
Please answer the questions below.
Svetllana [295]

Answer:

nice no questions

=)

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Identify the property being demonstrated
My name is Ann [436]

Answer:

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \dfrac{x}{5}  = 7 \\  \implies \: x = 7 \times 5 \\  \implies \: x = 35

So,b. multiplication

3 0
3 years ago
Read 2 more answers
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