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3 years ago

Instructions: Factor 715 as a product of primes.

Mathematics

1 answer:

Drupady [299]3 years ago

8 0

Answer:

{1, 5, 143}

Step-by-step explanation:

715 is evenly divisible by 5, so one factor of 715 is 5 and the next is 715/5, or 143. 143 is not factorable further. Thus, the factors of 715, in ascending order, are 1, 5 and 143.

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Which of the following is not a function?

stepladder [879]

Answer:

Step-by-step explanation:

d is not a function because if you do the vertical test for d

the line goes through it twice and if it goes through it twice it means it's not a function, D did not pass the vertical test

I put two pictures so you can see the difference. it shows you if it passed the vertical tesft or not. if you do not know abt the vertical test, you will learn abt it soon. a vertical test is very helpful to find out if it is a function or not. THE VERTICAL TEST CAN ONLY BE USED ON GRAPHS nothing more. I hope I helped.

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3 years ago

What is the ratio then what is the value of the ration?

Alekssandra [29.7K]

Answer:

For example, if there are eight oranges and six lemons in a bowl of fruit, then the ratio of oranges to lemons is eight to six (that is, 8∶6, which is equivalent to the ratio 4∶3).

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2 years ago

For the hypothesis test H0: μ = 10 against H1: μ >10 and variance known, calculate the Pvalue for each of the following test

Basile [38]

Answer:

a) $p_v =P(Z>2.05)=1-P(z$

b) $p_v =P(Z>-1.84)=1-P(z$

c) $p_v =P(Z>0.4)=1-P(z$

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Hypothesis

Null hypothesis: $\mu=10$

Alternative hypothesis: $\mu >10$

If the random variable is distributed like this: $X \sim N(\mu,\sigma)$

We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:

$z_o=\frac{\bar X -\mu}{\sigma}$

Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:

Part a

$p_v =P(Z>2.05)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(2.05,0,1,TRUE)"

Part b

$p_v =P(Z>-1.84)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(-1.84,0,1,TRUE)"

Part c

$p_v =P(Z>0.4)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(0.4,0,1,TRUE)"

Conclusions

If we use a reference value for the significance, let's say $\alpha=0.05$ . For part a the $p_v$ so then we can reject the null hypothesis at this significance level.