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tensa zangetsu [6.8K]
3 years ago
13

Does this table represent a function ?why or why not ?

Mathematics
1 answer:
Jlenok [28]3 years ago
8 0
Yes c. that is the correct answer.
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Solve for z <br><br> please help ?!???? or explain?
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Answer:

Step-by-step explanation:

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Please help this question is confusing
Nuetrik [128]

12, 15, 18

Step-by-step explanation:

The sign means equal to or greater than.

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Which option shows 18 divided by 15 as a fraction or mixed number
frutty [35]
\dfrac{18}{15} =  1\dfrac{18-15}{15} =  1\dfrac{3}{15}
6 0
3 years ago
Question: 5 ( x +20)= 7x +30
mel-nik [20]

Answer:35

Simplifying

5(x + 20) = 7x + 30

Reorder the terms:

5(20 + x) = 7x + 30

(20 * 5 + x * 5) = 7x + 30

(100 + 5x) = 7x + 30

Reorder the terms:

100 + 5x = 30 + 7x

Solving

100 + 5x = 30 + 7x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-7x' to each side of the equation.

100 + 5x + -7x = 30 + 7x + -7x

Combine like terms: 5x + -7x = -2x

100 + -2x = 30 + 7x + -7x

Combine like terms: 7x + -7x = 0

100 + -2x = 30 + 0

100 + -2x = 30

Add '-100' to each side of the equation.

100 + -100 + -2x = 30 + -100

Combine like terms: 100 + -100 = 0

0 + -2x = 30 + -100

-2x = 30 + -100

Combine like terms: 30 + -100 = -70

-2x = -70

Divide each side by '-2'.

x = 35

Simplifying

x = 35

Hope it helps. Please tell me if im correct

7 0
3 years ago
Read 2 more answers
At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 519 ml. The filling process follows a n
k0ka [10]

Answer:

1) The normal distribution should be used for the sample mean because the population distribution is known to be normal (answer b).

2) C. H0: mu = 519, H_1: mu not equal to 519, reject if z> 1.96 or z< -1.96.

3) Yes. There is enough evidence to support the claim that the true mean is not 519 ml.

Step-by-step explanation:

1) When the population follows a normal distribution, it is correct to assume a normal distribution for the sample mean.

2) As it is a two-tailed decision rule, we are interested in detecting a significant difference below and above the mean. This is why we use the unequal sign in the alternative hypothesis.

The null hypothesis state that there is not significant difference from 519.

The critical value for a significance level of 5% is z=1.96.

H_0: \mu=519\\\\H_a:\mu\neq 519

3) The claim is that the true mean is not 519 ml.

Then, the null and alternative hypothesis are:

H_0: \mu=519\\\\H_a:\mu\neq 519

The significance level is 0.05.

The sample has a size n=16.

The sample mean is M=522.

The standard deviation of the population is known and has a value of σ=6.

We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{16}}=1.5

Then, we can calculate the z-statistic as:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{522-519}{1.5}=\dfrac{3}{1.5}=2

This test is a two-tailed test, so the P-value for this test is calculated as:

\text{P-value}=2\cdot P(z>2)=0.046

As the P-value (0.046) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the true mean is not 519 ml.

7 0
3 years ago
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