All categories

3 years ago

PLS HELPPPP MEEE (explain ur answer)

Mathematics

1 answer:

adell [148]3 years ago

7 0

Answer:

(4). is the correct answer

Step-by-step explanation:

(- 3, 1, 1)

(1). - 3 - 1 - 1 ≠ - 2 not an answer...

(2). - 3 - 1 + 3(1) ≠ 2 not an answer...

(3). 4( - 3) + 1 - 3(1) ≠ - 4 not an answer...

(4). 2(- 3) + 3(1) + 4(1) = 1 

3(1) + 3(1) = 6 

- 3 = - 3(1) 

You might be interested in

What must be true about the average rate of change between any two points on the graph of an increasing function?

frez [133]

Are there options to choose from?

4 0

3 years ago

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

3 0

3 years ago

Brad has two lengths of copper pipe to fit together one has a length of 2 5/12 feet and the other has a length of 3 7 12 ft how

leva [86]

To find the total length of the two copper pipes, we need to add them together.

2 5/12 + 3 7/12 = 5 12/12

Since 12/12 = 1 or a whole, we can simplify this further.

5 12/12 = 6

Hope this helps!

4 0

3 years ago

How to graph this slop (put a pic if u can )pls help due at 11

poizon [28]

Answer:

hopes this help luv <3

Step-by-step explanation: