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3 years ago

Solve this equation 3/7x+5=-4/7x-3

Mathematics

2 answers:

ratelena [41]3 years ago

8 0

Answer:

x = -8

Step-by-step explanation:

$\frac{3}{7} x + 5 = \frac{-4}{7} x - 3\\5 = -1x-3\\8 = -1x\\-8 = x$

Alika [10]3 years ago

7 0

The answer is x= -1/8

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It is C trust me I’m doing the same thing

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3 years ago

Is Johhny is 91 years old and jac is 9 years old add the years togeter what is it?

leva [86]

Answer:

100 years old

Step-by-step explanation:hope it helps:)

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3 years ago

Read 2 more answers

Q 4.19: To verify whether there are negative consequences of taking a new type of medicine, 3000 tests were conducted. Assume th

tiny-mole [99]

Answer:

The p-value obtained is less than the significance level at which the test was performed at, hence, we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the the new medicine is potentially harmful.

Step-by-step explanation:

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, the null hypothesis is that the new medicine has no negative effects and the alternative hypothesis is that the new medicine is potentially harmful

Mathematically,

The null hypothesis is represented as

H₀: p = 0

The alternative hypothesis is represented as

Hₐ: p > 0

To do this test, we will use the z-distribution because although no information on the population standard deviation is known, the sample size is large enough.

So, we compute the test statistic

z = (x - μ)/σₓ

x = sample proportion = (31/3000) = 0.0103

μ = p₀ = 0

σₓ = standard error = √[p(1-p)/n]

where n = Sample size = 3000

σₓ = √[0.0103×0.9897/3000] = 0.0018462936 = 0.0018463

z = (0.0103 - 0) ÷ 0.0018463

z = 5.60

checking the tables for the p-value of this test statistic

Significance level = 1% = 0.01

The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for z = 5.60, at 0.01 significance level, with a one tailed condition) = < 0.00001

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.01

p-value = < 0.00001

p-value < 0.00001 < 0.10

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the the new medicine is potentially harmful.

Hope this Helps!!!

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4 years ago

What is the image of point A after a rotation of -90 degrees about the origin?

Anestetic [448]

Answer:

Step-by-step explanation:

A rotation of -90 degrees about the origin is equivalent to a clockwise rotation. Point A, initially at (2, 0), would become point A', (0, -2).

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4 years ago

For each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts. A large shipmen

Stella [2.4K]

Answer:

(a) The p-value of the test statistic should be 0.33.

(b) No, there is not sufficient evidence to reject the entire shipment.

Step-by-step explanation:

We are given that for each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts.

A quality control manager randomly selects 50 of the parts from the shipment and finds that 6 parts are defective.

Let p = proportion of defective parts among the current shipment.

So, Null Hypothesis, $H_0$ : p $\leq$ 10% {means that the defective parts in the shipment is at most 10%}

Alternate Hypothesis, $H_A$ : p > 10% {means that the defective parts in the shipment is greater 10%}

The test statistics that would be used here One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of parts that are defective among the current shipment = $\frac{6}{50}$ = 12%

n = sample of parts from the shipment = 50

So, test statistics = $\frac{0.12-0.10}{\sqrt{\frac{0.12(1-0.12)}{50} } }$

= 0.44

The value of z test statistics is 0.44.

(a) Now, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.44) = 1 - P(Z $\leq$ 0.44)

= 1 - 0.67003 = 0.3299 ≈ 0.33

(b) Since, the P-value of test statistics is more than the level of significance as 0.33 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the defective parts in the shipment is at most 10%.

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3 years ago