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3 years ago

SOMEONE HELP ASAP PLEASE. Answers and links that have nothing to do with the question will be reported.

Mathematics

1 answer:

Serhud [2]3 years ago

5 0

Answer:

$\frac{1}{15}$

Step-by-step explanation:

There are 2 yellow marbles, 3 pink marbles, and 5 blue marbles in the bag, for a total of 10 marbles.

Let's start with the probability that the first marble drawn is pink. Since there are 3 pink marbles out of the 10 total marbles, the probability of this happening is $3/10$ .

Since the first marble is not put back in the bag, there are now 9 marbles left, 2 of then being pink. The chances of drawing a pink marble now is $2/9$ . To find the chances of both of these events happening, multiply them together to get:

$\frac{3}{10}\cdot \frac{2}{9}=\frac{6}{90}=\boxed{\frac{1}{15}}$

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Find the range of the function. Use interval notation. g(x)=log↓3(x-2)

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Answer:

10x=log6-3

Step-by-step explanation:

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3 years ago

While visiting a memorial, a person approximated the angle of elevation to the top of the memorial to be 35º. After walking 255

Alenkasestr [34]

Draw a diagram to illustrate the problem as shown in the figure below.

Let h the height of the hill. =

At position A, the angle of elevation is 40°, and the horizontal distance to the foot of the hill is x.

By definition,

tan(40°) = h/x h = x tan40 = 0.8391x

(1)

At position B, Joe is (x - 450) ft from the foot of the hill. His angle of elevation is

40 + 18 = 58°.

By definition, tan(58°) = h/(x - 450)

h = (x - 450) tan(58°) = 1.6003(x-450)

h = 1.6003x - 720.135 (2)

Equate (1) and (2).

1.6003x - 720.135 = 0.8391x 0.7612x = 720.135

x = 946.0523

From (1), obtain

h = 0.8391*946.0523 = 793.8 ft

Answer: The height of the hill is approximately 794 ft (nearest integer)

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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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4 years ago

Solve the system by elimination. First, clear the denominators.

NARA [144]

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3 years ago

2<br> — x = 6<br> 3<br><br> Any help? Please?

Mama L [17]

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3 years ago

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