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zubka84 [21]
3 years ago
8

Solve this for me: (-9) − 15

Mathematics
2 answers:
Yakvenalex [24]3 years ago
4 0
The answer would be -24
elena-s [515]3 years ago
3 0

Answer:

-24

Hope that this helps!

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Two digit-
igor_vitrenko [27]

Let the number = 10x+y

x+y = 9

y-x = 1

(x+y)+(y-x) = 9+1

2y = 10

y = 10/2

<h3><u>y = 5</u></h3>

x+y = 9

9-y = x

x = 9 - 5

<h3><u>x = 4</u></h3>

<h2>So, your number is <u>45</u></h2>

... Hope this will help...

4 0
4 years ago
What is the slope of the faster rider
s344n2d4d5 [400]

Answer:

I think it’s 192 bcuz that the higtest number there so I think it will go faster

Step-by-step explanation:

8 0
3 years ago
1/4 times the sum of a number and −3.2 is 1.8
Bogdan [553]

10 2/5 is the number......

4 0
3 years ago
Read 2 more answers
Write 0.00973 in scientific notation
12345 [234]
9.73, or 9.73x10^-3, hope this helped!
5 0
3 years ago
Complete the statement by filling in the blanks: When constructing a confidence interval, if the level of confidence increases,
Yuki888 [10]

Answer:

\hat p \pm z_{\alpha/2} SE_{\hat p}

And for this case the margin of error would be:

ME= z_{\alpha/2} SE_{\hat p}

If the level of confidence increase we can conclude that the value of z_{\alpha/2} would increase and the the confidence interval would be wider, since the margin of error increase.

c. Increase; wider

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Let's assume that we have a parameter of interest p and we want to estimate the value of p with \hat p and in general the confidence interval if the distribution of p is normal is given by:

\hat p \pm z_{\alpha/2} SE_{\hat p}

And for this case the margin of error would be:

ME= z_{\alpha/2} SE_{\hat p}

If the level of confidence increase we can conclude that the value of z_{\alpha/2} would increase and the the confidence interval would be wider, since the margin of error increase.

c. Increase; wider

8 0
3 years ago
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