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2 years ago

I need helpful answers pls

Mathematics

1 answer:

Ivenika [448]2 years ago

7 0

Answer:

X=5 the angle is 76 degrees

Step-by-step explanation:

The bottom angles are corresponding. So 52+52=104, so you need 76 more degrees. Then simple math to caculate x.

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West Virginia math glencone course 2 volume 1 page 191

Step2247 [10]

Answer:

Im not sure but I think its -5

7 0

2 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

2 years ago

Solve the equation. (3b - 4)/5 = -b +4

Kryger [21]

Answer:

b=3

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

b=3

6 0

3 years ago

1. The US Mint has a specification that pennies have a mean weight of 2.5g. From a sample of 37 pennies, the mean weight is foun

Evgesh-ka [11]

Answer:

the pennies does not conform to the US mints specification

Step-by-step explanation:

z = (variate -mean)/ standard deviation

z= 2.5 - 2.4991 / 0.01648 = 0.0546

we are going to check the value of z in the normal distribution table, which is the table bounded by z.

checking for z= 0.0 under 55 gives 0.0219 (value gotten from the table of normal distribution)

we subtract the value of z from 0.5 (1- (0.5+0.0219)) = 0.4781 > 0.05claim

since 0.4781 > 0.05claim, therefore, the pennies does not conform to the US mints specification

the claim state a 5% significance level whereas the calculated significance level is 47.81%. therefore, the claim should be rejected

7 0

3 years ago

Jacob made a circle-shaped poster for his geometry class.

krok68 [10]

<span>circumference is L=2 $\pi r$
L=2π9=18π=18*3.14=56.52 in
Answer: </span>circumference is <span>18π or 56.52 in</span>

5 0

3 years ago