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3 years ago

5

hello! So the people in Indian server were bad(as they would not answer and instead spam) so I switched servers and I hope you a

ll will help me! pleaase

1 answer:

Art [367]3 years ago

7 0

Answer:

$\displaystyle E=26log2-24log3+8log 5$

Step-by-step explanation:

<u>Logarithms</u>

There are certain properties of the logarithms we'll need to recall in order to simplify the expression given in the question. We'll list them below

$log(x.y)=logx+logy\\log(x/y)=logx-logy\\logx^y=y.logx$

We'll also need to factor the following numbers into their prime factors:

$20=2^2\cdot 5$

$16=2^4$

$15=3\cdot 5$

$25=5^2$

$24=2^3\cdot 3$

$81=3^4$

$80=2^4\cdot 5$

The original expression is

$\displaystyle E=log20+16log\frac{16}{15}+12log\frac{25}{24}+log\frac{81}{80}$

Plugging in the factored values

$\displaystyle E=log(2^2\cdot 5)+16log\frac{2^4}{3\cdot 5}+12log\frac{5^2}{2^3\cdot 3}+log\frac{3^4}{2^4\cdot 5}$

Applying the properties we have

$\displaystyle E=log2^2+log 5+16(log2^4-log3-log 5)+12(log5^2-log2^3-log 3)+log3^4-log2^4-log 5$

$\displaystyle E=2log2+log 5+16log2^4-16log3-16log 5+12log5^2-12log2^3-12log 3+log3^4-log2^4-log 5$

$\displaystyle E=2log2+log 5+64log2-16log3-16log 5+24log5-36log2-12log 3+4log3-4log2-log 5$

Simplifying

$\boxed{\displaystyle E=26log2-24log3+8log 5}$

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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 quadrilateral is dilated by a scale factor of 2. It's original side measures are 7, 14, 21, and 28.

algol13

Answer:

14,28,42 and 56

Step-by-step explanation:

Multiply all lengths by 2 as the quadrilateral is dilated by a scale factor of 2

6 0

2 years ago

Read 2 more answers

Dionne can fold 175 packing boxes in 50 mìnutes. Elias can fold 120 packing boxes in 40 minutes. How long will it take each pers

Gre4nikov [31]

<h2>Dionne will fold 210 packing boxes in "60 mìnutes" and</h2><h2>Elias will fold 210 packing boxes in "70 mìnutes".</h2>

Step-by-step explanation:

Given,

Dionne can fold 175 packing boxes = 50 mìnute and

Elias can fold 120 packing boxes = 40 minutes

To find, the total time each person to fold 210 packing boxes = ?

∵ Dionne can fold 175 packing boxes = 50 mìnute

∴ In 1 mìnute, Dionne can fold number of packing boxes = $\dfrac{175}{50}$ = 3.5

∴ Dionne can fold 210 packing boxes = $\dfrac{210}{3.5}$ minutes

= 60 minutes

Also,

Elias can fold 120 packing boxes = 40 minutes

∴ In 1 mìnute, Elias can fold number of packing boxes = $\dfrac{120}{40}$ = 3

∴ Elias can fold 210 packing boxes = $\dfrac{210}{3}$ minutes

= 70 minutes

Thus, Dionne will fold 210 packing boxes in "60 mìnutes" and

Elias will fold 210 packing boxes in "70 mìnutes".

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3 years ago

In the diagram EF is the bisector of <DEF = m < D' EF.

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hii! i just wanted to say hope you're having a fantastic day and hope that things are going well (: stay strong and stay safe!

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2 years ago