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2 years ago

Find the balance in the account. $800 principal earning 7%, compounded annually, after 4 years

2 answers:

8 0

Use the compound amount formula: A = P (1+r)^t

Here,

A = $800 (1+0.07)^4 = $800 ( 1.31) = $1048.64 (rounded off to the nearest cent)

scoray [572]2 years ago

6 0

800×(1.07)^(4) =1,048.63

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What values of b satisfy 3(2b + 3)2 = 36?

meriva

What values of b satisfy 3(2b+3)^2 = 36

we have
3(2b+3)^2 = 36
Divide both sides by 3
(2b+3)^2 = 12
take the square root of both sides
( 2b+3)} =(+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3

b1=\frac{\sqrt{12}}{2} -\frac{3}{2}
b1=\sqrt{3} -\frac{3}{2}

b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
therefore

the answer is
the values of b are
b1=\sqrt{3} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}

9 0

2 years ago

Fabio and carlos play on a basketball team together. in the last game, fabio had 777 points less than 222 times as many points a

jasenka [17]

For this case, the first thing we must do is define variables.

We have then:

f: number of points that fabio scored.

c: number of points that Carlos scored.

We now write the equation that models the problem:

$f = 2c - 7$

Then, for f = 31 we have:

$2c - 7 = 31$

From here, we clear the number of points:

$2c = 31 + 7$

$2c = 38$

$c = \frac{38}{2}$

$c = 19$

Answer:

The equation to find the number of carlos points is:

$2c - 7 = 31$

Then, Carlos scored:

$c = 19$

5 0

2 years ago

Read 2 more answers

A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a

BigorU [14]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: $\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246$

For point a:

hypotheses are:

$H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\$

degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha = 0.05$ , reject null hypothesis if $t$ .

test statistic:

$t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55$

because the $t=-2.553$ , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at $\alpha =0.01$ , removing the null hypothesis if $t$ .

because $t=-2.553 >-2.908$ , fail to removing the null hypothesis.

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

$H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5$

Degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha =0.05$ , removing the null hypothesis if $t$ .

test statistic: $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986$

Since $t-1.986 >-1.895$ , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.