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Ilya [14]

3 years ago

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Mathematics

1 answer:

Ostrovityanka [42]3 years ago

5 0

Answer:11

Step-by-step explanation:

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Marco leaves the bathroom faucet running for 22 minutes while he gets ready in the morning. running the faucet for 5 minutes was

lord [1]

The amount of energy waste, in terms of hours which a 60-watt light bulb would remain on, is 61.6 hours then the correct option is B.

<h3>What are ratio and proportion?</h3>

A ratio is an ordered couple of numbers a and b, written as a/b where b can not equal 0. A proportion is an equation in which two ratios are set equal to each other.

Marco leaves the bathroom faucet running for 22 minutes while he gets ready in the morning.

Running the faucet for 5 minutes wastes the same amount of energy as leaving a 60-watt light bulb on for 14 hours.

Then we have

$\begin{aligned} \dfrac{5\ minutes}{14 \ hours } &=\dfrac{1\ minute }{x \ hours}\\\\x &= 2.8 \ hours \end{aligned}$

So, the rate is 2.8 hours per minute.

Since macro leaves the bathroom faucet running for 22 minutes. Then the waste of energy is leaving a 60-watt light bulb will be

→22 minutes × 2.8 hours/minute = 61.6 hours.

More about the ratio and the proportion link is given below.

brainly.com/question/14335762

5 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

3 years ago

Write the equation using function notation f(x). then find f(0)<br><br> 2x^2 - 3y = 6

Grace [21]

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