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2 years ago

I will give brainliest!!!!

Mathematics

2 answers:

Pie2 years ago

7 0

Answer:

Step-by-step explanation:

<em><u>2k + 4h is the same thing as 4h + 2k.</u></em>

QveST [7]2 years ago

4 0

Answer:

The. Answer. C

Step-by-step explanation:

You might be interested in

The physician tells a woman who usually drinks 5 cups of coffee daily to cut down on her coffee consumption by 75%. If this woma

Maru [420]

Answer:

Step-by-step explanation:

5 cups per day.....1 cup = 8oz.....so 5 cups = (5 * 8) = 40 oz

cut down by 75%....means ur only drinking 25%

25% of 40 oz =

0.25(40) = 10 oz per day <===

6 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

1 year ago

In a sale, normal prices are reduced by 13%. The sale price of a DVD recorder is £108.75 Calculate the normal price of the DVD r

kifflom [539]

Answer:

$122.89

Step-by-step explanation:

The two number you have in this equation are 108.75 and 13% (or .13)

first you want to multiply 108.75 to .13

108.75 * .13 = 14.1375

Now, you add 14.1375 to 180.75 and round to the nearest hundredth

180.75+ 14.1375=122.8875