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kicyunya [14]
3 years ago
8

Question 3 of 10

Mathematics
1 answer:
dsp733 years ago
3 0

Answer:

D. Junk bonds are high-risk bonds.

Step-by-step explanation:

This is right for Ap ex

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Assume that the significance level is a=0.05. Use the given information to find the P-value and the critical value(s). The test
Minchanka [31]

Answer:

The required P value is 0.0901.

The critical value for the test statistic is 1.645.

Step-by-step explanation:

Consider the provided information.

The test statistic of z=1.34 is obtained when testing the claim that p> 0.1

It is given given that z=1.34 as the claim has greater than inequality so it is a right tailed test.

Part (A) P-value

P\ value=1-P(Z\leq z)\\P\ value=1-P(Z\leq 1.34)

From the Standard normal distribution table P(Z\leq 1.34) =0.9099.

Therefore,

P\ value=1-0.9099=0.0901

Hence, the required P value is 0.0901.

Part (B) The critical value(s) is/are z=

It is given that the significance level is α=0.05

Using standard z value table we get the critical value for the test statistic is 1.645.

Hence, the critical value for the test statistic is 1.645.

6 0
4 years ago
I need extra help please, -7y^2-2y^2+y^3y-2y+5y^3-2y
joja [24]

Answer:

How to solve your problem

−

7

2

−

2

2

+

3

−

2

+

5

3

−

2

-7y^{2}-2y^{2}+y^{3}y-2y+5y^{3}-2y

−7y2−2y2+y3y−2y+5y3−2y

Simplify

1

Combine exponents

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7

2

−

2

2

+

3

−

2

+

5

3

−

2

-7y^{2}-2y^{2}+{\color{#c92786}{y^{3}y}}-2y+5y^{3}-2y

−7y2−2y2+y3y−2y+5y3−2y

−

7

2

−

2

2

+

4

−

2

+

5

3

−

2

-7y^{2}-2y^{2}+{\color{#c92786}{y^{4}}}-2y+5y^{3}-2y

−7y2−2y2+y4−2y+5y3−2y

2

Combine like terms

−

7

2

−

2

2

+

4

−

2

+

5

3

−

2

{\color{#c92786}{-7y^{2}}}{\color{#c92786}{-2y^{2}}}+y^{4}-2y+5y^{3}-2y

−7y2−2y2+y4−2y+5y3−2y

−

9

2

+

4

−

2

+

5

3

−

2

{\color{#c92786}{-9y^{2}}}+y^{4}-2y+5y^{3}-2y

−9y2+y4−2y+5y3−2y

3

Combine like terms

−

9

2

+

4

−

2

+

5

3

−

2

-9y^{2}+y^{4}{\color{#c92786}{-2y}}+5y^{3}{\color{#c92786}{-2y}}

−9y2+y4−2y+5y3−2y

−

9

2

+

4

−

4

+

5

3

-9y^{2}+y^{4}{\color{#c92786}{-4y}}+5y^{3}

−9y2+y4−4y+5y3

4

Rearrange terms

−

9

2

+

4

−

4

+

5

3

{\color{#c92786}{-9y^{2}+y^{4}-4y+5y^{3}}}

−9y2+y4−4y+5y3

4

+

5

3

−

9

2

−

4

{\color{#c92786}{y^{4}+5y^{3}-9y^{2}-4y}}

y4+5y3−9y2−4y

Solution

4

+

5

3

−

9

2

−

4

3 0
3 years ago
The junior varsity football team score 23 points in last Saturday's game they score the combination of seven point touchdowns in
frutty [35]

Answer:

  2 touchdowns, 3 field goals

Step-by-step explanation:

The number of touchdowns cannot be more than 3, so it is relatively easy to find the solution by trial and error.

  23 is not divisible by 3, so 0 touchdowns is not a solution

  23 -7 = 16 is not divisible by 3, so 1 touchdown is not a solution

  23 -14 = 9 is divisible by 3, so 2 touchdowns and 3 field goals is a solution

  21 -21 = 2 is not divisible by 3, so there is only one solution.

7 0
3 years ago
Which line contains the pint (6-4) and has a slope of -5/6
natima [27]

Answer:

y = - 5/6 x + 1

Step-by-step explanation:

y = mx + b      m: slope    b: y intercept

point (6 , -4)        x = 6     y = -4

b = y - mx = -4 - ((-5/6) x 6) = - 4 + 5 = 1

equation: y = - 5/6 x + 1

7 0
3 years ago
What is 5.138 + 2.624 = ? (show ur work)
Mrac [35]

Answer:

7.762

Step-by-step explanation:

Brainliest!

3 0
2 years ago
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