What point on the unit circle corresponds to an angle of 173º

A store sold a certain brand of jeans for $38 one day, the store sold 6 pair of jeans of that brand. How much did the 6 pairs of

baherus [9]

6 pairs of jeans would cost $228.

4 0

1 year ago

Read 2 more answers

Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit

KIM [24]

<span>Since both functions follow the form of 2X + 3Y = c in general, where c is some constant, both functions will have the same slope as of -2/3. However, both graph will not have the same Y-intercept as the profits are different. Similarly, the X-intercept will also be different due to the profits.

I hope my answer has come to your help. Have a nice day ahead and may God bless you always!
</span>

8 0

1 year 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

1 year ago

Helpp!!!? need help quick ..

stealth61 [152]

Note that

n^m*n^r=n^(m+r)

2^5*2^x=2^(5+x)

the answer is D

7 0

1 year ago

James is the manager at an entertainment arena that draws an average 7,000 patrons per event. Each ticket taker can process 350

Elina [12.6K]

Answer:

No. James didn't have enough ticket takers to process the average number of patrons hat usually attend the events.

He need to hire 2 more ticket taker (i.e 20 ticket takers) in order to process the average number of patrons hat usually attend the events.

Step-by-step explanation:

Given:

Average amount of patron per event= 7000

Each ticket taker can process = 350

Number of ticket takers hired = 18

We need to find the whether he hire enough to process the average number of patrons that usually attend the events.

Solution:

We will find the Average amount of patron ticket takers can process.

Now we can say that

Average amount of patron ticket takers can process is equal to Each ticket taker can process multiplied by Number of ticket takers hired.

framing in equation form we get;

Average amount of patron ticket takers can process = $350 \times 18 = 6300\ patrons$

Hence With the required hires James cannot fully process the patrons usually attending the event.

To find how many ticket takers required we will divide average number of patrons with Each ticket taker can process.

framing in equation form we get;

ticket takers required = $\frac{7000}{350}=20$

hence In order t process the the average number of patrons that usually attend the events James will require 20 ticket takers.

6 0

1 year ago