three consecutive even integers in which 20 more than twice of the third is 28 more than the sum of the first two integers

A box contains 24 transistors,4 of which are defective. If 4 are sold at random,find the following probabilities. i. Exactly 2 a

zavuch27 [327]

SOLUTION

This is a binomial probability. For i, we will apply the Binomial probability formula

i. Exactly 2 are defective

Using the formula, we have

$\begin{gathered} P_x=^nC_x\left(p^x\right?\left(q^{n-x}\right) \\ Where\text{ } \\ P_x=binomial\text{ probability} \\ x=number\text{ of times for a specific outcome with n trials =2} \\ p=\text{ probability of success = }\frac{4}{24}=\frac{1}{6} \\ q=probability\text{ of failure =1-}\frac{1}{6}=\frac{5}{6} \\ ^nC_x=\text{ number of combinations = }^4C_2 \\ n=\text{ number of trials = 4} \end{gathered}$

Note that I made the probability of being defective as the probability of success = p

and probability of none defective as probability of failure = q

Exactly 2 are defective becomes the binomial probability

$\begin{gathered} P_x=^4C_2\times\lparen\frac{1}{6})^2\times\lparen\frac{5}{6})^{4-2} \\ P_x=6\times\frac{1}{36}\times\frac{25}{36} \\ P_x=\frac{25}{216} \\ =0.1157 \end{gathered}$

Hence the answer is 0.1157

(ii) None is defective becomes

$\begin{gathered} \lparen\frac{5}{6})^4=\frac{625}{1296} \\ =0.4823 \end{gathered}$

hence the answer is 0.4823

(iii) All are defective

$\begin{gathered} \lparen\frac{1}{6})^4=\frac{1}{1296} \\ =0.00077 \end{gathered}$

(iv) At least one is defective

This is 1 - probability that none is defective

$\begin{gathered} 1-\lparen\frac{5}{6})^4 \\ =1-\frac{625}{1296} \\ =\frac{671}{1296} \\ =0.5177 \end{gathered}$

Hence the answer is 0.5177

3 0

10 months ago

What is the slope of the line?

Ann [662]

Answer:

m= 8/5

Step-by-step explanation:

Points on the graph: ( -4, -4) and (1, 4)

Slope:

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

m=(4+4)/(1+4)

m= 8/5

4 0

2 years ago

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True or false square root 54 is close to 7?

antoniya [11.8K]

No it's over : 7.3 , unless your prompts context is close to as in getting there or close in general?

3 0

3 years ago

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GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

Now, P-value of the test statistics is given by;

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

2 years ago

One of Newton’s laws of motion states that objects in motion tend to stay in motion. What is the reason that this law may not al

Nastasia [14]

The answer is D.

The full specification of Newton's First Law is (from Wikipedia)

In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.

On earth we're generally in slightly a non-inertial frame, with rotation and revolution around the sun. It's also impossible to turn of gravity, so objects are being acted upon by an outside force. Typically there's also the outside force of friction, which makes things tend to stop.

4 0

2 years ago

Read 2 more answers