What percent is $15.60 of $156.00?

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

(c) 90.53%

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.

8 0

3 years ago

Longer congruent shorter proportional

irinina [24]

Shorter. Because of the way the arches work. the pink dot is the center.

8 0

3 years ago

1.

Harlamova29_29 [7]

Answer:

a) x = 1

Step-by-step explanation:

Given: 2x + 3 = 5

1. Subtract 3 from both sides:

2x + 3 - 3 = 5 - 3

⟹ 2x = 2

2. Divide both sides by 2:

2x ÷ 2 = 2 ÷ 2

⟹ x = 1

3. Check your work:

2(1) + 3 = 5

2 + 3 = 5

⟹ 5 = 5 ✔

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8 0

2 years ago

Solving the equation

liubo4ka [24]

Answer:

y = 5

Step-by-step explanation:

Step 1: Write equation

-(4y + 2) - (-3y - 6) = -1

Step 2: Solve for y

Distribute: -4y - 2 + 3y + 6 = -1
Combine like terms: -y + 4 = -1
Subtract 4 on both sides: -y = -5
Divide both sides by -1: y = 5

Step 3: Check

Plug in y to verify it's a solution.

-(4(5) + 2) - (-3(5) - 6) = -1

-(20 + 2) - (-15 - 6) = -1

-22 - (-21) = -1

-22 + 21 = -1

-1 = -1

7 0

3 years ago

1.3 divided by 0.0338 step by step please

kati45 [8]

Alright, so we have 1.3/0.0338. Since it's easier (in my opinion) to work with whole numbers, we can multiply the fraction by 10000/10000 to get 13000/338. With a bit of guess and check, we can see that
338*30=338*3*10
1 2 (what I carry is at the top)
338
x3
____
1114

Multiplying that by 10, I get 11140, which isn't enough. Trying 338*40, which is 338*4*10, we can add 338 to 338*3 to get 338*4 to get

2
1114
+338
____
1462

Multiplying that by 10, we get 14620, which is more than 13000 - something we don't want. Repeating this for 338*35 (which is 338*3.5*10, and 3.5 is 3*338+338/2)=11830 and which isn't enough, we then move on to something between 35 and 40 (the number doesn't matter), say 39. 338*39=338*3.9*10, and 338*3.9 is 338*3+338*9/10, and
338*39 results to 13182, which is more than 13000 , but only by a tiny bit, so we can try 38 using the same method, getting 12844, which is smaller, so we know it's between 38 and 39. Finding the difference between 13000 and 12844, we get 13000-12844=156 and the answer is therefore 38+156/338

4 0

3 years ago