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icang [17]
3 years ago
11

Help what's the answer to this??

Mathematics
2 answers:
Evgesh-ka [11]3 years ago
7 0

Answer:

I think it's J

Step-by-step explanation:

I may be wrong but I hope this helps :)

Degger [83]3 years ago
3 0

Answer:

J

Step-by-step explanation:

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(h+1)^2=16 solve using square root property
IrinaVladis [17]

Answer: h = 3

Step-by-step explanation:

(h+1)² = 16

h+1 = √16

h+1 = 4

h = 3

Hope this helped :)

3 0
2 years ago
Read 2 more answers
What is the 45th digit in the decimal equivalent of 1/7
Luden [163]
1/7<span> =  0.</span><span>142857 since this decimal is repeating divide 45/6(number of repeating numbers) and it gives you 7.5 so you know that it repeats this sequence 7 times plus an extra 3 numbers so the 45th decimal is 2</span>
3 0
3 years ago
8+2[10(7x2-9)-40] simplify
soldier1979 [14.2K]

Answer:

140x2−252

Step-by-step explanation:

7 0
3 years ago
Italian delight sells three sizes of pizzas at different prices. If you buy all three, it costs a total of $46.24. A medium pizz
beks73 [17]

Answer:

$12.99

Step-by-step explanation:

This is because 15.75+17.5= 33.25. 46.24-33.25= 12.99 and therefore, a small pizza should cost 12.99.

4 0
4 years ago
A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on th
Lunna [17]

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.  

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value for pretest , y = test value for posttest  

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90  

The system of hypothesis for this case are:  

Null hypothesis: \mu_D \geq 0  

Alternative hypothesis: \mu_D < 0  

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

The first step is calculate the difference d_i=x_i-y_i and we obtain this:  

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833  

The third step would be calculate the standard deviation for the differences, and we got:  

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763  

The 4 step is calculate the statistic given by :  

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494  

The next step is calculate the degrees of freedom given by:  

df=n-1=6-1=5  

Now we can calculate the p value, since we have a left tailed test the p value is given by:  

p_v =P(t_{(5)}  

The p value is lower than the significance level assumed \alpha=0.05, so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest \mu_{pretest}-\mu_{postest} are less than 0.

3 0
3 years ago
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