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3 years ago

What is the Sector area with radius 6 and an angle of 50°? Use 3.14 for pi.

Mathematics

1 answer:

nevsk [136]3 years ago

7 0

Answer:

15.7 units²

Step-by-step explanation:

The sector area is 50/360 times the area of the circle (since the central area of a complete circle is 360° and that for the sector is 50°), or

Sector area = (50/360)(3.14)(6 units)² = 15.7 units²

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What is the rate of change of a line that passes through the points (-3,5) and (2,-1)?

lara [203]

Answer:

3/4

Step-by-step explanation:

The slope of the line is

m = (y2-y1)/(x2-x1)

= (-2--5)/(6-2)

= (-2+5)/(6-2)

=3/4

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3 years ago

Tayler has $320 to pay for dining room chairs. She expects to pay about $80 per chair. Her friend told her that she has 3 that T

VARVARA [1.3K]

Answer:

320/80 +3 = c

Step-by-step explanation:

if she has $320 and each chair is $80, then dividing 320 by 80 will tell you how many chairs she can buy (4), plus 3 because she gets 3 chairs for free. it equals c because c is the total amount of chairs she gets

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3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

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3 years ago

Fx) = 3x+2<br> What is f(5)?

Alexeev081 [22]

Answer:

the correct answer is 17

4 0

3 years ago

1.Mr. Bowen’s test is normally distributed with a mean of 75 and a standard deviation of 3 points.

lianna [129]

Answer:

2.28%

Step-by-step explanation:

Mr. bowens test is normally distributed with a mean (μ) of 75 and a standard deviation (σ) of 3 points.

The z score is used in probability to show how many standard deviation is a raw score below or above the mean. The formula for the z score (z) is given by:

$z=\frac{x-\mu}{\sigma}$

For a raw score (x) of 81 points, the z score can be calculated by:

$z=\frac{x-\mu}{\sigma}=\frac{81-75}{3}=2$

Therefore from the normal probability distribution table, the probability that a randomly selected score is greater than 81 can be given as:

P(x > 81) = P(z > 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228 = 2.28%

7 0

3 years ago

What is the Sector area with radius 6 and an angle of 50°? Use 3.14 for pi.​

What is the Sector area with radius 6 and an angle of 50°? Use 3.14 for pi.