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Lorico [155]
2 years ago
11

PLEASE SHOW ALL WORK

Mathematics
1 answer:
EleoNora [17]2 years ago
5 0

Answer:

t = 2,3

Step-by-step explanation:

s (t) = t^2 + 2 t + 5

p (t) = 11 + 3 t

A:

s (1) = 8

s (2) = 13

s (3) = 20

s (4) = 29

Now

t^2 + 2 t + 5 = 11 t + 3t \\\\t^2 - t - 6 = 0\\\\t^2 - 3 t - 2 t - 6 = 0 \\\\(t-2)(t-3)=0\\\\t = 2, 3

(b) The values are same at t =2 an t =3.  

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Which of the following expressions are equivalent to 6-3?
Zigmanuir [339]

Answer:

1/216 and 1/6 raised to the 3 power.

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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].

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Answer:

See attached pictures.

Step-by-step explanation:

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The cosine function always starts and ends at the y=A.

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The amplitude is 5.

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8.) Here A = 2 so the amplitude is A. B is 2π so the period is 1. C is 1 so the graph is shifted up a unit.

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