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kumpel [21]
3 years ago
14

Can someone also help me on this one!!

Mathematics
1 answer:
alexdok [17]3 years ago
5 0

The change in the X values is in multiples of 2.

The change in the h(x) values need to be: -0.3 x 2 = -0.6

Now find the h(x) values that have a difference of -0.6

A negative value is a decrease.

2 to 4 is an increase.

4 to 6 is an increase.

6 to 8 is an increase.

8 to 10 is an increase.

10 to 12 = 20-19.8 = 0.2

12 to 14 = 19.8 - 19.2 = 0.6

The two columns are 12 and 14

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Help plz. I'll mark brainliest
abruzzese [7]
I answered this question just now! Did you add another question like this?

1. rule = r - 12 (number = 25)
2. rule = a + 34 (number = 43) 
3. rule = s - 5 (number = 20)
4. rule = b + 16 (number = 32)
5. rule = w + 3 (number = 9)
6. rule = n + 9 (number = 42)

7. 15 - n when n = 9 would be 15 - 9, and your answer would be 6.

8. "32 more than a number d" implies add d to 32, so your answer is C. 32 + d
6 0
3 years ago
Item to ship 58 , <br> items per case 6
9966 [12]

Answer: 9 and 2/3 cases  

Step-by-step explanation:

6 goes into 58 9 times to make 54 so makes 4 left over 4/6 simplified makes 2/3.

6 0
1 year ago
Read 2 more answers
One step equations x/4 =9
Tamiku [17]

Hi Natalie

We have x/4 = 9

All you have to do is cross multiplication

x = 4*9

x= 36

That's all :)

I hope that's help:)


Now if you wanna check my answer just replace x by 36

36/4 = 9

9= 9

Because when you divide 36 by 4 you will find 9


I hope that's help:)

8 0
3 years ago
Read 2 more answers
Compare 2/5 and 1/4
elena-14-01-66 [18.8K]

Answer:

2/5 > 1/4

Step-by-step explanation:

2/5 = 4/10 = 0.4

1/4 = 0.25

0.4 > 0.25

2/5 > 1/4

8 0
3 years ago
Read 2 more answers
Consider independent simple random samples that are taken to test the difference between the means of two populations. The varia
Arturiano [62]

Answer:

d. t distribution with df = 80

Step-by-step explanation:

Assuming this problem:

Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:

a. t distribution with df = 82.

b. t distribution with df = 81.

c. t distribution with df = 41.

d. t distribution with df = 80

Solution to the problem

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

This last one is an unbiased estimator of the common variance \sigma^2

So on this case the degrees of freedom are given by:

df= 37+45-2=80

And the best answer is:

d. t distribution with df = 80

5 0
2 years ago
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