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vovangra [49]
3 years ago
12

Solve the attachment...​

Mathematics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer:

2 ( Option A )

Step-by-step explanation:

The given integral to us is ,

\longrightarrow \displaystyle \int_0^1 5x \sqrt{x}\ dx

Here 5 is a constant so it can come out . So that,

\longrightarrow \displaystyle I =  5 \int_0^1 x \sqrt{x}\ dx

Now we can write √x as ,

\longrightarrow I = \displaystyle 5 \int_0^1 x . x^{\frac{1}{2}} \ dx

Simplify ,

\longrightarrow I =  5 \displaystyle \int_0^1 x^{\frac{3}{2}}\ dx

By Power rule , the integral of x^3/2 wrt x is , 2/5x^5/2 . Therefore ,

\longrightarrow I = 5 \bigg( \dfrac{2}{5} x^{\frac{5}{2}} \bigg] ^1_0 \bigg)

On simplifying we will get ,

\longrightarrow \underline{\underline{ I = 2 }}

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The area of triangle BXA is 15*sqrt(2).

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6+2(7×-3) Simplify the expression
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=36

Step-by-step explanation:

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A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 110, p = 0.2

So

\mu = E(X) = np = 110*0.2 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 22}{4.1952}

Z = 0.715

Z = 0.715 has a pvalue of 0.7626.

X = 17

Z = \frac{X - \mu}{\sigma}

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

4 0
3 years ago
15-2x=-7x what does x equal
Degger [83]

Answer:

x = -3

Step-by-step explanation:

15 - 2x = -7x

add 2x

15 = -5x

divide by -5

-3 = x

4 0
3 years ago
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