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4 years ago

Y=-16t^2 + 400. Solve for t

Mathematics

1 answer:

zavuch27 [327]4 years ago

8 0

Answer:

± 1/4 sqrt((400-y)) =t

Step-by-step explanation:

Y=-16t^2 + 400

Subtract 400 from each side

y - 400 = -16t^2 +400-400

y-400 = -16t^2

Divide by -16t^2

(y-400)/-16 = t^2

(400-y)/16 = t^2

Take the square root of each side

±sqrt((400-y)/16) =sqrt(t^2)

±sqrt((400-y)) / sqrt(16) =t

± 1/4 sqrt((400-y)) =t

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Simplify each expression 35x+63y

Murrr4er [49]

Find the Greatest Common Factor (GCF)

GCF = 7

Factor out the GCF (Write the GCF first and then in parenthesis, divide each term by the GCF)

7(35x/7 + 63y/7)

Simplify each term in parenthesis

7(5x + 9y)

6 0

3 years ago

Read 2 more answers

Use the t-distribution to find a confidence interval for a difference in means μ1-μ2 given the relevant sample results. Give the

Serggg [28]

Answer:

(a) The best estimate of $\mu_{1}-\mu_{2}$ is 13.2.

(b) The margin of error is 5.30.

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means using a t-interval is:

$CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

(a)

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean $\bar x$ is a point-estimate of the population mean μ.

Similarly the point estimate of the difference between two means is:

$\bar x_{1}-\bar x_{2}$

Compute the point estimate of $\mu_{1}-\mu_{2}$ as follows:

$E(\mu_{1}-\mu_{2})=\bar x_{1}-\bar x_{2}\\=79.0-65.8\\=13.2$

Thus, the best estimate of $\mu_{1}-\mu_{2}$ is 13.2.

(b)

Compute the pooled variance as follows:

$S_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}=\frac{(35-1)10.5^{2}+(20-1)7.2^{2}}{35+20-2}=89.311$

Compute the critical value of t as follows:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (35+20-2)}=t_{0.025, 53}=2.00$

*Use a t-table.

Compute the margin of error as follows:

$MOE=t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

$=2.00\times\sqrt{89.311\times [\frac{1}{35}+\frac{1}{20}]} \\=5.298\\\approx5.30$

Thus, the margin of error is 5.30.

(c)

Compute the 95% confidence interval for the difference between two means as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm MOE$

$=13.2\pm 5.298\\=(7.902, 18.498)\\\approx (7.90, 18.50)$

Thus, the 95% confidence interval for the difference between two means is (7.90, 18.50).

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4 years ago

given that the value of circumference of a circle is 2/3 of its area. determine the length of the radius of the circle!

sineoko [7]

$\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}$

Let's assume the redius of the given circle is (r).

so,

circumference = $\sf{ 2\pi r }$
Area = $\sf{ \pi r^{2} }$

According to the question, 

circumference of a circle is 2/3 of its area.

$:\implies \dfrac{2}{3}×Area=circumference\\\\:\implies \dfrac{2}{3}×\pi r^{2}=2\pi r\\\\:\implies \pi r^{2}=2\pi r×\dfrac{3}{2}\\\\:\implies \pi r^{2}=3\pi r\\\\:\implies 3=\dfrac{\pi r^{2}}{\pi r}\\\\ :\dashrightarrow r=3$