All categories

3 years ago

Help asap!!!

1 answer:

3 0

$\bf \textit{arc's length}\\\\ s=\cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\[-0.5em] \hrulefill\\ r=6.5\\ s=5.7 \end{cases}\implies 5.7=\cfrac{\theta \pi (6.5)}{180}\implies 1026=6.5\pi \theta \\\\\\ \cfrac{1026}{6.5\pi }=\theta \implies \stackrel{\textit{rounded up, using }\pi =3.14}{50.27=\theta }$

You might be interested in

Point B is on line segment AC. If AB = x + 6, BC = x + 8 and AC = 10, then find the value of x.

kumpel [21]

Step-by-step explanation:

this simply means that the sum of both short segments must be the same as the length of the whole line.

so,

x + 6 + x + 8 = 10

2x + 14 = 10

x + 7 = 5

x = 5 - 7 = -2

7 0

2 years ago

What is the estimate of 19.7-6.9

Zolol [24]

Always to do each to 1 sigfig 20-6=14

5 0

4 years ago

Please help on problem 14! Thanks!!

pickupchik [31]

2 hours has more the the other so I would say it would be 8

4 0

3 years ago

A small lawnmower company produced 1,500 lawnmowers in 2008. In an effort to determine how maintenance-free these units were, th

ikadub [295]

Answer:

The 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the average using the finite correction factor is:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

The information provided is:

$N=1500\\n=183\\\sigma=1.47\\\bar x=3.3$

The critical value of z for 95% confidence level is,

z = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

$=3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)$

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

7 0

3 years ago

An art history professor assigns letter grades on a test according to the following scheme.A: Top 5% of scoresB: Scores below th

adoni [48]

Answer:

Grade B score:

$76 \leq x \leq 89$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73.3

Standard Deviation, σ = 9.7

We are given that the distribution of score on test is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

B: Scores below the top 5% and above the bottom 62%

We have to find the value of x such that the probability is 0.62

$P( X > x) = P( z > \displaystyle\frac{x - 73.3}{9.7})=0.62$

$= 1 -P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.62$

$=P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.38$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 73.3}{9.7} = 0.305\\\\x = 76.26$

We have to find the value of x such that the probability is 0.05

$P(X < 0.95) = \\\\P( X < x) = P( z < \displaystyle\frac{x - 73.3}{9.7})=0.95$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 73.3}{9.7} = 1.645\\\\x = 89.26$

Thus, the numerical value of score to achieve grade B is

$76 \leq x \leq 89$

7 0

3 years ago