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tensa zangetsu [6.8K]

3 years ago

12

PLZ ANSWER ASAP A bag contains 6 blue marbles, 10 red marbles, and 9 green marbles. If two marbles are drawn at random without r

eplacement, what is the probability that two red marbles are drawn?

2 answers:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

10/25

10+9+6 equals 25 and there is 10 red marbles. I think this is the answer don't take my word for it.

svp [43]3 years ago

8 0

Answer:

I would determindedy say your answer is 3/25.

Step-by-step explanation:

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3(5x + 7) = 30 + 15x how many solutions

Xelga [282]

Answer:

No solution

Step-by-step explanation:

3(5x+7) = 30 + 15x

DISTRIBUTE

15x+21 = 30+15x

15x-15x = 0

So no solution

8 0

2 years ago

Read 2 more answers

A bag holds 11 beads 7 of the beads are red the rest are white two beads are taken at random work out the probability that one o

Anvisha [2.4K]

Answer:

28/55

Step-by-step explanation:

total no of beads are 11

red beads are 7

and white beads are 4

no of ways to take 2 beads from 11 beads are 11c2

that is( 11*10)/2=55

no of ways to take 1 red bead from 7 is 7c1 that is 7

no of ways to take 1 white bead from 4 is 4c1 is that is 4

required probability is( 7*4)/55=28/55

this question has already been answered on someone else's account so know this answer is 100% correct

hope that helped :)

3 0

2 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

(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

A figure is rotated and reflected. what can u say about the new figure in relation to the original figure?

mylen [45]

The new figure is formed differently and is not the same as the other ome

8 0

3 years ago

What is a solution and what is not?

Ksenya-84 [330]

Cant see the rest of the question ;-;

3 0

3 years ago