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3 years ago

What is the rate of change in the equation y=8x+6?

Mathematics

1 answer:

schepotkina [342]3 years ago

6 0

The rate of change is the number next to the x when in this form. Therefore the rate of change is 8

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6x-4(3x-5)=2 solve for x.

jeka57 [31]

Answer:

x = 3

Step-by-step explanation:

6x -4(3x-5) = 2

6x - 12x + 20 = 2

-6x = 2 - 20

-6x = -18

x = 3

5 0

3 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

Rewrite 1/4 1/3as a unit rate

noname [10]

1/4 = 0.25 = quarter

1/3 = 0.(3) = one third

3 0

3 years ago

You did 5 trials of push-ups in a minute with the following results:

Finger [1]

Answer:

36 push-ups on your sixth trial

<u>I would appreciate Brainliest, but no worries.</u>

8 0

3 years ago

I REALLY NEED HELP!

Snowcat [4.5K]

Answer:

Step-by-step explanation:

The compound interest formula is $A = P (1+r/n)^nt$ where:

P is the starting amount called the principle
r is the rat written as a decimal
n is the number of times compounded in a year
t is the number of years

Substitute a value into each variable to solve.

P = $147 since 10% of 1,470 is being invested which makes P = 0.10(1470) = 147.

The rate is 3.5% or r = 0.035.
n = 12 because it is compounded monthly meaning 12 times a year.
t = 25 since it will earn for 25 years.

$A = P (1+r/n)^{nt}\\A = 147(1 + \frac{0.035}{12})^{12*25}\\A = 147 ( 1 + 0.002916)^{300}\\A = 147(1.002916)^{300}\\A = 352.19$

Repeat this process for each formula.

5 0

4 years ago