All categories

3 years ago

Find the length a. a=?

Mathematics

1 answer:

algol [13]3 years ago

6 0

Cosine law
a²=b²+c²-2ab*cosA

a²=13²+11²-2*13*11*cos(108)≈378.37
a≈19.45

You might be interested in

I need help w/ this please help me!!

artcher [175]

Answer:B

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

1. (5pt) In a recent New York City marathon, 25,221 men finished and 253 dropped out. Also,

gladu [14]

Answer:

(a) Explained below.

(b) The 99% confidence interval for the difference between proportions is (-0.00094, 0.00494).

Step-by-step explanation:

The information provided is:

n (Men who finished the marathon) = 25,221

n (Women who finished the marathon) = 12,883

Compute the proportion of men and women who finished the marathon as follows:

$\hat p_{1}=\frac{25221}{25221 +253}=0.99\\\\\hat p_{2}=\frac{12883 }{12883+163}=0.988$

The combined proportion is:

$\hat P=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}\\\\=\frac{25221+12883}{38520}\\\\=0.989$

(a)

The hypothesis is:

H₀: The rate of those who finish the marathon is the same for men and women, i.e. p₁ - p₂ = 0.

Hₐ: The rate of those who finish the marathon is not same for men and women, i.e. p₁ - p₂ ≠ 0.

Compute the test statistic as follows:

$Z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)\cdot [\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

$=\frac{0.99-0.988}{\sqrt{0.989(1-0.989)\times[\frac{1}{25474}+\frac{1}{13046}]}}\\\\=1.78$

Compute the p-value as follows:

$p-value=2\times P(Z>1.78)\\\\=2\times 0.03754\\\\=0.07508\\\\\approx 0.075$

The p-value of the test is more than the significance level. The null hypothesis was failed to be rejected.

Thus, concluding that the rate of those who finish is the same for men and women.

(b)

Compute the 99% confidence interval for the difference between proportions as follows:

The critical value of z for 99% confidence level is 2.58.

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.99-0.988)\pm 2.58\times\sqrt{\frac{0.99(1-0.99)}{25474}+\frac{0.988(1-0.988)}{13046}}\\\\=0.002\pm 0.00294\\\\=(-0.00094, 0.00494)\\\\$

Thu, the 99% confidence interval for the difference between proportions is (-0.00094, 0.00494).

6 0

3 years ago

2. Express the given statement as an algebraic expression. Sum ofa and 5 divided by 8

bezimeni [28]

Answer:

An algebraic expression for Sum of a and 5 divided by 8 is $\mathbf{\frac{a+5}{8}}$

Step-by-step explanation:

We need to express the given statement as an algebraic expression.

Sum of a and 5 divided by 8

Basically, we are given the expression in English sentence and we have to write algebraic expression.

It can be done in two steps:

Sum of a and 5 = We know that sum is denoted by + sign. So, we get (a+5)

Sum of a and 5 divided by 8 = $\frac{a+5}{8}$

So, An algebraic expression for Sum of a and 5 divided by 8 is $\mathbf{\frac{a+5}{8}}$

7 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :