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bixtya [17]
3 years ago
8

Pete has 5 marbles . Jay has a number of marbles that is two more than 5 . How many marbles does jay have?

Mathematics
2 answers:
Sliva [168]3 years ago
8 0

Answer:

The answer is 7

Step-by-step explanation:

kogti [31]3 years ago
5 0

Answer:

7 marbles

Step-by-step explanation:

You would put five first because that's how much marbles pete had and for jay you just add a 2.

5+2=7

Hope this helps:)

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Write the equation of the line perpendicular to 2x - 6y = 12 that passes through the point (-3,0).
Trava [24]

Answer:

y = -3x -9

Step-by-step explanation:

slope = 1/3

perpendicular slope = -3

y = mx + b

0 = -3(-3) + b

-9 = b

y = -3x -9

7 0
2 years ago
Read 2 more answers
The model represents an equation. What value of x makes the equation true?
Slav-nsk [51]

Answer:

A) 29/8

Step-by-step explanation:

On the left side of the equals sign, we have five x's and nine -1's.

On the right side of the equals sign, we have three -x's and twenty 1's.

Both sides are equal, so:

5(x) + 9(-1) = 3(-x) + 20(1)

5x − 9 = -3x + 20

Add 3x to both sides.

8x − 9 = 20

Add 9 to both sides.

8x = 29

Divide both sides by 8.

x = 29/8

5 0
3 years ago
Please help me with the answer
liraira [26]

Answer:

-18

Step-by-step explanation:

-15+6=-9

the absolute value of -9 is 9

and 9*-2=-18

5 0
1 year ago
Read 2 more answers
Please find the distance between the points
marusya05 [52]

Answer:

they are 6 uints apart

Step-by-step explanation:

5 0
3 years ago
The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of the girls
Elanso [62]

Answer:

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=22 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =12.9^2=166.41 represent the sample variance obtained

\sigma^2_0 =16.1^2 =259.21 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis: \sigma^2 \geq 259.21

Alternative hypothesis: \sigma^2

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

3 0
3 years ago
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