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tester [92]
3 years ago
15

Will give brainly if right

Mathematics
1 answer:
garri49 [273]3 years ago
8 0

Answer:

you are correct, it is all of the above

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What is the equation of the line parallel to the given line with an X intercept of 4
mixas84 [53]

You haven't shared the given line, so all I can do here is to invent a line and then show you how to write the equation of a new line which is parallel to mine and which has an x-intercept of 4.


My invented line: y = (2/3)x + 3


The new line MUST have the same slope: m = 2/3.


Then y = mx + b becomes y = (2/3)x + b. Find the x-intercept by setting y = 0 and solving for x: (2/3)x = 0 - b. Now replace x with 4 and find b:


-b = (2/3)(4) = 8/3. Then b = -8/3, and the new line is


y = (2/3)x - 8/3.

3 0
3 years ago
Read 2 more answers
Express {x+1}{2x-3} as a trinomial.
Marizza181 [45]

Answer:

2x^2 + 5x + 3

Step-by-step explanation:

First, expand the original equation:

(x + 1)(2x + 3)

2x^2 + 3x + 2x + 3

Now simplify:

2x^2 + 5x + 3

And that's your answer!

6 0
3 years ago
What is 1/15 as a decimal
ElenaW [278]
1/15 as a decimal would be 0.6666666......

Or you can round that off to 0.67 



7 0
3 years ago
Read 2 more answers
Solve b <br> 3b / 7-1 = 5
ruslelena [56]

Answer:

b= 2 over 7

Step-by-step explanation:

exact form b= 2 over 7

decimal form b= 0.285714

7 0
3 years ago
Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
Vesnalui [34]

Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 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.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so 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".  

Calculate the statistic

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

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

Since is a one left tailed test the p value would be:  

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0
3 years ago
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