All categories

3 years ago

Solve for x and y 2x+7=3y-1

1 answer:

7 0

To solve for x and y, isolate each variable and simplify your equation:

Solve for x:
2x+7=3y-1 (Subtract 7 on each side)
2x=3y-8 (Divide by 2 on each side)
x= $\frac{3y}{2} -4$

Solve for y:
2x+7=3y-1
2x+8=3y (Add 1 on each side)
$\frac{2x}{3} + \frac{8}{3}$ =y (Divide by 3)
y= $\frac{2x}{3} + \frac{8}{3}$

~I hope this helps!~

You might be interested in

What is the given value of the variable 7/8 times C for C equals eight

goldenfox [79]

7/8 * C

C = 8

7/8 * 8

56/8

7

6 0

3 years ago

Please help <3

Aleksandr [31]

Answer:

x =3

Step-by-step explanation:

3(4x + 4) = 2(5x+9) - 12

First step is to open the brackets by multiplying. It becomes

3×4x + 3×4 = 2×5x + 2×9 -12

12x + 12 = 10x + 18 - 12

Collecting like terms on the right hand side and left hand side of the equation, it becomes

12x -10x = 18 -12 -12

2x = -6

Negative sign on the right hand side of the equation overwhelms the positive sign on left hand side of the equation. It becomes

2x = -6

x = -6/2 = -3

Checking

3(4×-3 + 4) = 2(5×-3+9) - 12

-24 = -12-12= -24

So x = -3

3 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

Help on this question<br>

AnnyKZ [126]

Answer:

20(pie)

Step-by-step explanation:

Since the circumference of a circle is 20 feet, the formula for circumference is 2r(pie) with r=radius

(radius)2=diameter

(diameter)(pie)

20(pie)

6 0