1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
arlik [135]
4 years ago
9

Translate the phrase into an algebraic expression. The product of 5 and x

Mathematics
1 answer:
LekaFEV [45]4 years ago
5 0

Answer:

5x.

Step-by-step explanation:

The product of 5 and x is the same thing as asking what is 5 times x.

That is 5 * x = 5x.

Hope this helps!

You might be interested in
Anton has a deck that is 890 cm by 2891 cm. If he wants to add 66 cm. how large would his deck be?
Aliun [14]

Answer:

The answer is 2,763,796 cm². Or 276.3796m², which is the footprint of a nice size house.

Step-by-step explanation:

890cm \times 2891cm = 2572990 {cm}^{2}

890 + 66 = 956

956cm \times 2891cm = 2763796 {cm}^{2}

4 0
3 years ago
Hours
REY [17]
$5 per hour is the constant rate because 10/2 is 5, 20/4 is 5 etc.
5 0
3 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
The wool of a vicuna is finer than any other wool. The hair of this animal can be as thin as 0.000006 meters. If the average thi
lorasvet [3.4K]

Answer:

7.8 × 10-5 meters

Step-by-step explanation:

the length of the a human hair= 8.4 x 10^-5

length of the vicuna hair:0.000006 = 0.6 x 10^-5

===========================================

difference between human hair and vicuna hair:

8.4 x 10^-5 - 0.6 x 10^-5 = 10^-5 ( 8.4 - 0.6)

= 10^-5 (7.8)

= 7.8 x 10 ^-5 meters

6 0
3 years ago
Read 2 more answers
If a snail travels 200 inches in 2 hours, how long will it take for the snail to travel 50 inches?
iren2701 [21]
It would take it 30 minutes.
4 0
4 years ago
Read 2 more answers
Other questions:
  • If you double the height of a cone, what does it do to its volume?
    8·1 answer
  • Which of the following formulas would find the surface are of a right cylinder where h is the height, r is the radius, and BA is
    14·1 answer
  • If QRST below is a rectangle, what is the measure of angle R + angle S?
    13·1 answer
  • Calculate the distance penny her bicycle if he row she rides 1/4 mile each day for 5 day.
    5·1 answer
  • 17 - (9 + 10) = (17 - 9) + 10 A.True B.False
    9·2 answers
  • What is the line of symmetry for y=2x^2+3x+1
    12·2 answers
  • What is the answer K/2+1-1k=-2k
    6·1 answer
  • What is the area of these I will mark as brainliest
    10·1 answer
  • 12. The roots of a quadratic are x = -5 and x = 3.
    15·1 answer
  • Write the number from least to greatest 0.6 1/2 2/3 0.39
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!