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4 years ago

Two congruent squares are joined to make a rectangle. The perimeter of the rectangle is 96 cm.

2 answers:

4 0

Answer is C.
288 cm2

----------------------------------------
96 / 2 = 48

each square has side = 12

so new rectangle has L = 24, W = 12

A = 12 x 24 = 288

loris [4]4 years ago

3 0

D.
144 cm2
Two congruent squares are joined to make a rectangle. The perimeter of the rectangle is 96 cm.

What is the area of the rectangle?

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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.

8 0

3 years ago

BRAINLIEST + Points! Help in need and please explain!

Elanso [62]

C., because 60 is half of 120, and 12% is half of 24%

hope it is right

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3 years ago

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Alex owns a miniature model of a train. The actual train is 14 feet tall and 70 feet long. If Alex's model is 6 inches tall, wha

rjkz [21]

6 is multiplied by 2 1/3 to get to 14. 70 divided by 2 1/3 is 30. The model of the train is 30 inches long.

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2 years ago

In the 1920s, the continued rise in the stock market and economic growth depended most on

Pachacha [2.7K]

Credit. That's why the stock market crash was so terrible in 1929.

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3 years ago

QUICK !! What is the image point of (2,-7) after a translation right 4 units and up 3 units ?

Paul [167]

Answer:

(6, - 4)

Step-by-step explanation:

A translation of 4 units right is equivalent to adding 4 to the x- coordinate

A translation of 3 units up is equivalent to adding 3 to the y- coordinate

(x, y) → (x + 4, y + 3) ← translation rule

Hence

(2, - 7) → (2 + 4, - 7 + 3) → (6, - 4)

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3 years ago

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