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Andrew [12]
3 years ago
15

Factor by grouping.......

Mathematics
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

A. (7n-1) (8n-1)

Step-by-step explanation:

56n^2 -8n-7n+1

56n^2 -8n       -7n+1

Factor out 8n from the first group   and -1 from the second group

8n (7n-1)     -1(7n-1)

Factor out the (7n-1)

(7n-1) (8n-1)

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A student in wildlife management studied trout habitat in the upper Shavers Fork watershed in West Virginia. The springtime wate
nataly862011 [7]

Answer:

C. H 0 : μ = 7 vs. Ha : μ ≠ 7

Since the calculated value of t = -9.462 falls in the critical region  t ≤-2.048

We conclude that the  springtime water the tributary water basin around the Shavers Fork watershed is not neutral. We accept our alternate hypothesis and reject the null hypothesis.

Step-by-step explanation:

The null hypothesis the usually the test to be performed. Here we want to check whether the water is neutral or not. Neutral water must have a pH of 7 . This can be stated as the null hypothesis. And the claim is treated as the alternate hypothesis that water in not neutral or not having pH= 7

In symbols it will be written as

H0: : μ = 7 vs. Ha : μ ≠ 7

So choice C is the best option for this hypothesis testing.

Let the significance level be 0.05

The degrees of freedom = n-1= 29-1 = 28

The critical value is  t ≥ 2.048 and t ≤ - 2.048 for 0.05 two tailed test with 28 df.

The test statistic to use is t- test

t= x- u/ s/√n

The total sum is 170.9 and mean = x= 5.893

The u = 7

And the sample standard deviation is =s= 0.63

Putting the values

t= 5.893-7/0.63/√29

t= - 1.107/0.11699

t= -9.4623

Since the calculated value of t = -9.462 falls in the critical region  t ≤-2.048

We conclude that the  springtime water the tributary water basin around the Shavers Fork watershed is not neutral. We accept our alternate hypothesis and reject the null hypothesis.

4 0
3 years ago
How to solve <br> 12 x 3/4
tamaranim1 [39]

Answer:

9

Step-by-step explanation:

Write the fractions in fraction form 12/1 x 3/4

Multiply numerator x numerator and denominatior x denominato

You will get 36/4

36/4 = 9

5 0
3 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
3 years ago
Solve the two-step equation. 5.1 = –3x – 4.2
Tems11 [23]
5.1 = -3x - 4.2 

9.3=-3x 

-3.1=x 


3 0
4 years ago
Read 2 more answers
What is the solution to this system of linear equations? 12q + 3r = 15 –4q – 4r = –44 (–18, 29) (–2, 13) (8, –1) (15, –44)
beks73 [17]

Answer:

(-2,13)

I believe this is the answer.

8 0
3 years ago
Read 2 more answers
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