Please answer whoever gets it right brainliest

Abigail and her children went into a movie theater and she bought $71.50 worth of

snow_lady [41]

Answer: The number of drinks she bought would be <u><em>6</em></u> and the number of pretzels she bought would be <u><em>10. </em></u>

6 0

3 years ago

You invested $500 in a savings account and after three years you had $650 in your bank. What was the interest rate? Use the equa

SpyIntel [72]

Answer:

An increase of 43.3% per year

Step-by-step explanation:

using the equation, plug in the respective variables. 650 =500R3, 3 times 500 is 1500. 650=1,500R. divide both sides by 1500 to get .4333, 3 repeating. .433 in percent form is 43.3%. (I just use 650 out of 500 or 650/500 which is 1.3 divided by 3 because 3 years to get the same amount)

6 0

3 years ago

one number is three more than four time another if the sum of the two number is 23 then find both numbers

zloy xaker [14]

<h3><u>The value of x is 19.</u></h3><h3><u>The value of y is 4.</u></h3>

x = 3 + 4y

x + y = 23

Because we have a value for x we can plug it directly into the second equation.

3 + 4y + y = 23

Subtract 3 from both sides.

4y + y = 20

Combine like terms.

5y = 20

Divide both sides by 5.

y = 4

Plug this value for x back into the original equation.

x = 3 + 4(4)

x = 3 + 16

x = 19

8 0

3 years ago

Given two independent random samples with the following results: n1=8x‾1=186s1=33 n2=7x‾2=171s2=23 Use this data to find the 90%

KonstantinChe [14]

Answer:

The point of estimate for the true difference would be:

$186-171= 15$

And the confidence interval is given by:

$(186-171) -1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= -10.753$

$(186-171) +1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= 40.753$

Step-by-step explanation:

For this case we have the following info given:

$\bar X_1 = 186$ the sample mean for the first sample

$\bar X_2 = 171$ the sample mean for the second sample

$s_1 =33$ the sample deviation for the first sample

$s_2 =23$ the sample deviation for the second sample

$n_1 = 8$ the sample size for the first group

$n_2 = 7$ the sample size for the second group

The confidence interval for the true difference is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

We can find the degrees of freedom are given:

$df = n_1 +n_2 -2 =8+7-2= 13$

The confidence level is given by 90% so then the significance would be $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ we can find the critical value with the degrees of freedom given and we got: