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1 year ago

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The volume of air inside a rubber ball with radius r can be found using the function V(r) = four-thirds pi r cubed. What does V

and five-sevenths represent?
the radius of the rubber ball when the volume equals five-sevenths cubic feet
the volume of the rubber ball when the radius equals five-sevenths feet
that the volume of the rubber ball is 5 cubic feet when the radius is 7 feet
that the volume of the rubber ball is 7 cubic feet when the radius is 5 feet

2 answers:

enot [183]1 year ago

5 0

Answer:

5/7

Step-by-step explanation:

the volume of the rubber ball when the radius equals five-sevenths feet

Nat2105 [25]1 year ago

4 0

Corrected Question

The volume of air inside a rubber ball with radius r can be found using the function V(r) = four-thirds pi r cubed. What does V(5/7) represent?

Answer:

(B)the volume of the rubber ball when the radius equals five-sevenths feet

Step-by-step explanation:

The Volume of a sphere of radius r can be found using the formula:

$V(r)=\dfrac43 \pi r^3$

Therefore comparing the expression: $V(\dfrac57)$ with V(r):

$Radius, r=\dfrac57$

Thus, $V(\dfrac57)$ is the volume of a ball of radius $\dfrac57$ feet.

The correct option is B.

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Factor the polynomial expression 15x^2 - 2x - 8

AVprozaik [17]

Answer:

(3 x + 2) (5 x - 4)

Step-by-step explanation:

Factor the following:

15 x^2 - 2 x - 8

Factor the quadratic 15 x^2 - 2 x - 8. The coefficient of x^2 is 15 and the constant term is -8. The product of 15 and -8 is -120. The factors of -120 which sum to -2 are 10 and -12. So 15 x^2 - 2 x - 8 = 15 x^2 - 12 x + 10 x - 8 = 5 x (3 x + 2) - 4 (3 x + 2):

5 x (3 x + 2) - 4 (3 x + 2)

Factor 3 x + 2 from 5 x (3 x + 2) - 4 (3 x + 2):

Answer: (3 x + 2) (5 x - 4)

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2 years ago

To buy a computer, Raquel borrowed $3,000 at 9%

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Answer:

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Step-by-step explanation:

We have the amount she will pay back, but first, we need to find the Interest accrued.

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$I = \frac{P * R * T }{100}$

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T = time taken (in years)

Therefore, the interest on $3,000 at 9% simple interest for 4 years is:

$I = \frac{3000 * 9 * 4}{100}$

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$3000 + $1080 = $4080

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This year, a hardware store has a profit of -$6.0 million. This profit is 3/4 of last year's profit. What is last year's profit?

katen-ka-za [31]

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2 years ago

which shows the correct substitution of the values a, b, and c from the equation 0 = 4x2 2x – 1 into the quadratic formula below

irinina [24]

$0 = 4x^2+2x- 1$

Quadratic formula is

$x = \frac{-b+-\sqrt{b^2-4ac}}{2a}$

'a' is the coefficient of x^2 = 4

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Now we plug in all the values in quadratic formula

$x = \frac{-b+-\sqrt{b^2-4ac}}{2a}$

$x = \frac{-2+-\sqrt{2^2-4(4)(-1)}}{2(4)}$

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1 year ago

Read 2 more answers

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

lisov135 [29]

Answer:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

$\bar X_1= 75.1$ represent the sample mean for the scores of the undergraduate students

$s_1 = 12.8$ represent the standard deviation for the undergraduate students

$n_1 =35$ the sample size for the undergraduate

$\bar X_2= 72.1$ represent the sample mean for the scores of the high school students

$s_2 = 14.6$ represent the standard deviation for the high school students

$n_2 =50$ the sample size for the high school

The confidence interval for the true difference of means is given by:

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

The degrees of freedom are given by:

$df=n_1 +n_2 -2= 35+50-2=83$

The confidence level is 90% and the significance level is $\alpha=0.1$ and $\alpha/2 =0.05$ then the critical value would be:

$t_{\alpha/2}= 1.99$

And replacing the info we got:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

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1 year ago