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3 years ago

2(3x - 1) + 2(4x + 5) = 8

Mathematics

2 answers:

professor190 [17]3 years ago

7 0

Distribute as we get,

$[(3x*2)-(2*1)]+[(4x*2)+(5*2)]=8\\(6x-2)+(8x+10)=8$

Remove the brackets,

$6x-2+8x+10=8$

Then combine,

$14x+8=8$

Move +8 to the right side and change the +8 to -8 (The operator must be the opposite when move to the other side.)

$14x=8-8\\14x=0$

Move 14 to the right side so we can find the value of x. (Because it's x multiply 14 then 0 will be divided by 14.)

$x=\frac{0}{14}$

Therefore, x = 0.

tigry1 [53]3 years ago

3 0

Answer:

x=0

Step-by-step explanation:

2(3x - 1) + 2(4x + 5) = 8

Distribute

6x -2 +8x +10 =8

Combine like terms

14x +8 = 8

Subtract 8 from each side

14x+8-8=8-8

14x =0

Divide each side by 14

14x/14 = 0/14

x=0

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We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

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Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

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