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gladu [14]
3 years ago
5

Please answer whoever gets it right brainliest

Mathematics
1 answer:
lubasha [3.4K]3 years ago
7 0
2 less than n is n-2
And
n plus n is 2n
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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:

t_{\alpha/2}= \pm 1.77

The point of estimate for the true difference would be:

186-171= 15

And replacing into the formula for the confidence interval we got:

(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

5 0
3 years ago
Can someone please help me with this question! btw thx in advance. Use the remainder Theorem to find the remainder: (x^3-1)/(x-4
olga55 [171]
F(x)/(x-a)=f(a)
Here a=4
So we need f(4)
F(4)=(4)^3 -1
F(4)=64-1=63
5 0
3 years ago
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