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2 years ago

PLEASE HELP ME RUNNING OUT OF TIME!!!

Mathematics

1 answer:

elena-14-01-66 [18.8K]2 years ago

8 0

Answer:

I am not sure of the answer but it is between A and C but my final answer is gonna be C.

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Help me out here with this another equation the the sign on the left is the checking

omeli [17]

Put the same variable on the same side first than solve

5 0

2 years ago

A density graph is used to find the probability of a discrete random variable

IRINA_888 [86]

Answer:

False

Step-by-step explanation:

This is false because A density graph is not used to find the probability of a discrete random variable taking on a range of values. This is because you have to use a calculations instead of a graph. The correct how to calculate is: Determine a single event with a single outcome. Identify the total number of outcomes that can occur. Divide the number of events by the number of possible outcomes.

Therefore, it's B ( false).

4 0

3 years ago

What is the solution set for. <br> |X+3| = 5?

kotegsom [21]

The answer will be x = 2

8 0

3 years ago

Read 2 more answers

ANSWER ASAP PLSS AND HURRY

Rina8888 [55]

Answer:

not sure what it is....

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A professor finds that the average SAT score among all students attending his college is 1150 ± 150 (μ ± σ). He polls his class

solniwko [45]

Answer:

Step-by-step explanation

Hello!

Be X: SAT scores of students attending college.

The population mean is μ= 1150 and the standard deviation σ= 150

The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.

If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:

H₀: μ = 1150

H₁: μ ≠ 1150

α: 0.05

$Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)$

$Z_{H_0}= \frac{1200-1150}{\frac{150}{\sqrt{25} } } = 1.67$

The p-value for this test is 0.0949

Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.

I hope it helps!

7 0

3 years ago