Ask question

Ask question

All categories

3 years ago

5

The price, e, of an entertainment system at extreme electronics is $220 less than twice the price, u, of the same system at ultr

a electronics. the difference in price between the system at extreme electronics and ultra electronics is $175.
which system of linear equations can be used to determine the price of the system at each store?

1 answer:

MrMuchimi3 years ago

3 0

Answer: The price of system at extreme electronics is $570 and the price of system at ultra electronics is $395.

Step-by-step explanation:

Let the price of the system at the ultra electronics be u

Let the price of the system at extreme electronics be e

According to question, we have

$2u-e=220\\\\and\\\\e-u=175$

Now, using the substitution method, we will solve the above system of equations.

$e-u=175\\\\e=175+u$

Now, put the value of u in the first equation :

$2u-e=220\\\\2u-(175+u)=220\\\\2u-175-u=220\\\\u-175=220\\\\u=220+175\\\\u=\$395$

Now, we put the value of u in the equation which is given by:

$e=175+u\\\\e=175+395\\\\e=\$570$

Hence, the price of system at extreme electronics is $570 and the price of system at ultra electronics is $395.

You might be interested in

Question 10 of 24

skad [1K]

The answer would be: B

4 0

3 years ago

What is the variance of 4,7,5,9? how do i find the answer?

kumpel [21]

See attachment for math work and answer.

6 0

3 years ago

Mary spent 3/4 of her money on a jacket. She spent 1/4 of the remainder on a shirt. The jacket cost $33 more than the shirt. How

Leviafan [203]

$49.50 the jacket costs this because her total amount was $66. to find total money I said 3/4x - 1/4x = $33 so 1/2x =$33 and x= $66. then multiply the total by 3/4 to get $49.50

6 0

3 years ago

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

<em>Let </em> $\mu_1$ <em> = mean number of times men order take-out dinners in a month.</em>

<em /> $\mu_2$ <em> = mean number of times women order take-out dinners in a month</em>

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviation;

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 mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, <u>test statistics</u> = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = <u>0.0662</u>

4 0

3 years ago

A basketball coach is purchasing 12 shirts for her team, each with a different number. At the checkout counter, the clerk places

ICE Princess25 [194]

Kidjov the answer I came up with for you is 792 I hope this helps

4 0

3 years ago

Read 2 more answers