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Aleksandr-060686 [28]
2 years ago
12

3×=21, the value of ×is ​

Mathematics
1 answer:
lord [1]2 years ago
8 0

<em>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</em>

<em>➷  </em>x + 3 = 21<em />

<em>➷ </em>x = 21 - 3<em />

<em>➷ </em>x = 18<em />

<em>✽</em>

<em>➶ Hope This Helps You!</em>

<em>➶ Good Luck (:</em>

<em>➶ Have A Great Day ^-^</em>

<em>↬ May ♡</em>

<em />

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Deffense [45]
She should ask her parents how it works before doingt it

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3 years ago
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In triangle FGH, GJ is an angle bisector of ∠G and perpendicular to FH.
Citrus2011 [14]
The picture in the attached figure

we know that
If line GJ bisects the angle FGH, that means we also know that the angle FGH is divided into two equal sizes.

Now, we have
Angle FGJ = Angle HGJ
Angle GJF = Angle GJH
This information means that angle GHJ = angle GFJ

We have two congruent right-angled triangle
We can then further deduce that
Side GF = Side GH
Since triangle FGJ and triangle GJH are congruent,
then
side FJ equals to side JH
3x-8=16
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4 0
3 years ago
Find the scale factor of the similar shapes below.
timofeeve [1]

Answer:

x=12 and the scale factor is 1.5

Step-by-step explanation:

From D' to F' is 7.5 and from D to F is 5cm the difference is 2.5. From F' to E' is x and from F to E is 8. This is just understanding what you can see. You find x by figuring out the ratio from 5 to 7.5 is and 8 to x or vise versa.  It depends which way you are going. I think you are going from E,D,F to E',D',F'. So from 5cm to 7.5 cm. This is 1.5 because you are dividing 7.5/5. If you needed to find x, it would be 12.

5 0
2 years ago
To test Upper H 0​: muequals50 versus Upper H 1​: muless than50​, a random sample of size nequals23 is obtained from a populatio
natta225 [31]

Answer:

Step-by-step explanation:

Hello!

1)

<em>To test H0: u= 50 versus H1= u < 50, a random sample size of n = 23 is obtained from a population that is known to be normally distributed. Complete parts A through D. </em>

<em> A) If  ¯ x = 47.9  and s=11.9, compute the test statistic .</em>

For thistest the corresponsing statistis is a one sample t-test

t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }~~t_{n-1}

t_{H_0}= \frac{47.9-50}{\frac{11.9}{\sqrt{23} } } = -0.846= -0.85

B) If the researcher decides to test this hypothesis at the a=0.1 level of significance, determine the critical value(s).

This test is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of the statistic. The ejection region is defined by one critical value:

t_{n-1;\alpha }= t_{22;0.1}= -1.321

Check the second attachment. The first row shows α= Level of significance; the First column shows ν= sample size.

The t-table shows the values of the statistic for the right tail. P(tₙ≥α)

But keep in mind that this distribution is centered in zero, meaning that the right and left tails are numerically equal, only the sign changes. Since in this example the rejection region is one-tailed to the left, the critical value is negative.

C) What does the distribution graph appear like?

Attachment.

D) Will the researcher reject the null hypothesis?

As said, the rejection region is one-tailed to the right, so the decision rule is:

If t_{H_0} ≤ -1.321, reject the null hypothesis.

If t_{H_0} > -1.321, do not reject the null hypothesis.

t_{H_0}= -0.85, the decision is to not reject the null hypothesis.

2)

To test H0​: μ=100 versus H1​:≠​100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts​ (a)-(d).

a) If x =104.2 and s=9.6, compute the test statistic.

For this example you have to use a one sample t-test too. The formula of the statistic is the same:

t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{104.2-100}{\frac{9.6}{\sqrt{24} } = } = 2.143

b) If the researcher decides to test this hypothesis at the α=0.01 level of​ significance, determine the critical values.

This hypothesis pair leads to a two-tailed rejection region, meaning, you'll reject the null hypothesis at either small or big values of the statistic. Then the rejection region is divided into two and determined by two critical values (the left one will be negative and the right one will be positive but the module of both values will be equal).

t_{n-1;\alpha/2 }= t_{23; 0.005}= -2.807

t_{n-1;1-\alpha /2}= t_{23;0.995}= 2.807

c) Draw a​ t-distribution that depicts the critical​ region(s). Which of the following graphs shows the critical​ region(s) in the​t-distribution?

Attachment.

​(d) Will the researcher reject the null​ hypothesis?

The decision rule for the two-tailed hypotheses pair is:

If t_{H_0} ≤ -2.807 or if t_{H_0} ≥ 2.807, reject the null hypothesis.

If -2.807 < t_{H_0} < 2.807, do not reject the null hypothesis.

t_{H_0}= 2.143 is greater than the right critical value, the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the test statistic is not between the critical values.

3)

Full text in attachment. The sample size is different by 2 but it should serve as a good example.

H₀: μ = 20

H₁: μ < 20

a) n= 18, X[bar]= 18.3, S= 4, Compute statistic.

t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }= \frac{18.3-20}{\frac{4}{\sqrt{18} } } = -1.80

b) The rejection region in this example is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of t.

Out of the three graphics, the correct one is A.

c)

To resolve this you have to look for the values in the t-table that are the closest to the calculated t_{H_0}

Symbolically:

t_{n-1;\alpha_1 } \leq t_{H_0}\leq t_{n-1;\alpha _2}

t_{H_0}= -1.80

t_{17; 0.025 }= -2.110

t_{17;0.05}= -1.740

Roughly defined you can say that the p-value is the probability of obtaining the value of t_{H_0}, symbolically: P(t₁₇≤-1.80)

Under the distribution the calculated statistic is between the values of -2.110 and -1.740, then the p-value will be between their cumulated probabilities:

A. 0.025 < p-value < 0.05

d. The researcher decides to test the hypothesis using a significance level of α: 0.05

Using the p-value approach the decision rule is the following:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

We already established in item c) that the p-value is less than 0.05, so the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the p-value is less than α.

I hope this helps!

6 0
3 years ago
Solve the equation for m:<br><br> m/5 = 60
alex41 [277]

Answer:

m = 300

Step-by-step explanation:

m/5 = 60

multiply both sides of the equation by 5

m = 60 * 5 = 300

3 0
2 years ago
Read 2 more answers
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