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Oliga [24]
3 years ago
14

(b) About what percentage of the states (out of 51) seem to have close to 800 licensed drivers per 1000 residents? (Round your a

nswer to one decimal place.)
Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0
This question uses a dotplot graph from which you have to extract the information.

I have attached the dotplot graph in a pdf file. You can open the file to see the graph if you do not have it.

The graph shows with dots (they use this mark +) the 51 states. Every dot means that one state has the number of license drivers per 1000 residents given by the line. For example, you can see that just above the mark of 700 therer are two dots, meaning that two states have 700 licensed drivers per 1000 residents.

Now you can see that there are 5 dots "close" the 800 mark.

That means that 5 out of 51 states "seem to have close to 800 licensed drivers per 1000 residents".

The percentage is calculated as [5/51]*100 = 9.804 %, which rounded to one decimal place is 9.8%.

Answer: 9.8%

 
Download pdf
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The triangles below are similar.<br> Find the value of x
11111nata11111 [884]

Answer:

23

Step-by-step explanation:

91/70=(29+x)/40

5 0
3 years ago
I will upvote and give brainliest answer
Citrus2011 [14]

Answer:

x = 58 degrees

Step-by-step explanation:

Angle K (which is x) and Angle I are congruent corresponding angles with one another, so if, as an example, Angle I was 10, Angle K would also be 10.

<u><em>All you need to do in this case is add Angle H (56) and Angle A (66) together:</em></u>

56 + 66 = 122

<u><em>Then subtract 122 from 180 (because triangles add up to 180 degrees) to get x:</em></u>

180 - 122 = 58

So, x = 58 degrees.

7 0
3 years ago
Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard devi
vladimir1956 [14]

Answer:

(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.

(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.

(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.

(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.

The significance level for the test is <em>α</em> = 0.05.

The decision rule is:

If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 11.

The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.

The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₀ > 1.91) = 0.043.

The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>

The sample size is, <em>n</em> = 17.

The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.

The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>

The sample size is, <em>n</em> = 7.

The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.

The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is not rejected at 5% level of significance.

(d)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 28.

The test statistic value is, <em>t</em> = 2.13 ≈ 2.12.

The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₂₇ > 2.12) = 0.022.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

5 0
3 years ago
The difference of the degrees of the polynomials p(x) = 3x2y2 + 5xy? - x6 and q(x) = 3x5 - 4x3 + 2 is
svet-max [94.6K]

Answer:

The difference of the degrees of the polynomials p (x) and q (x) is 1.

Step-by-step explanation:

A polynomial function is made up of two or more algebraic terms, such as p (x), p (x, y) or p (x, y, z) and so on.

The polynomial’s degree is the highest exponent or power of the variable in the polynomial function.

The polynomials provided are:

p(x) = 3x^{2}y^{2} + 5xy - x^{6}\\\\q(x) = 3x^{5} - 4x^{3} + 2

The degree of polynomial p (x) is:

\text{deg}\ p (x)=6

The degree of polynomial q (x) is:

\text{deg}\ q (x)=5

The difference of the degrees of the polynomials p (x) and q (x) is:

\text{deg}\ p(x)-\text{deg}\ q(x)=6-5=1

Thus, the difference of the degrees of the polynomials p (x) and q (x) is 1.

6 0
3 years ago
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AlekseyPX

Answer:

Step-by-step explanation:

Hope this helps :)

3 0
3 years ago
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