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amid [387]
3 years ago
12

the claim that 20% of americans know their credit score is tested. suppose that the sample evidence causes us to fail to reject

the null hypothesis claim that 20% of americans know their credit score. state the final conclusion in simple non technical terms
Mathematics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

Conclusion : People ≠ 20% don't know about their credit score.

Step-by-step explanation:

Hypothesis is testing a statement for its statistical significance.

Null Hypothesis (H0)  implies 'no difference from tested value', Alternate hypothesis (H1)  implies 'significant difference from tested value'

Let % of people knowing their credit score = CS

H0 : CS = 20

H1 : CS  ≠  20

If the null hypothesis is rejected, it implies that we reject the claim that CS i.e '% of americans knowing their credit score = 20%'. So, the alternate hypothesis is accepted, i.e we conclude that '% americans knowing their credit score ≠ 20%'.

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Integral 1+cos8x/tan2x-cot2x
UkoKoshka [18]
I believe it's<span> 8cos(x)⁸ - 16cos(x)⁶ + 10cos(x)⁴ - 2cos(x)².

</span>8cosx^8 - 16cosx^6+10cosx^4-2cosx^2<span>

Alternately, you can write [</span><span><span>1 / (tan(2x) - cot(2x))] + [cos(8x) / (tan(2x) - cot(2x))].

</span></span>\dfrac{1}{tan(2x)-cot(2x)}+ \dfrac{cos(8x)}{tan(2x)-cot(2x)}<span><span>
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5 0
3 years ago
A tunnel is built in form of a parabola. The width at the base of tunnel is 7 m. On
anastassius [24]

Given:

The width at the base of parabolic tunnel is 7 m.

The ceiling 3 m from each end of the base there are light fixtures.

The height to light fixtures is 4 m.

To find:

Whether it is possible a trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel.

Solution:

The width at the base of tunnel is 7 m.

Let the graph of the parabola intersect the x-axis at x=0 and x=7. It means x and (x-7) are the factors of the height function.

The function of height is:

h(x)=ax(x-7)             ...(i)

Where, a is a constant.

The ceiling 3 m from each end of the base there are light fixtures and the height to light fixtures is 4 m. It means the graph of height function passes through the point (3,4).

Putting x=3 and h(x)=4 in (i), we get

4=a(3)((3)-7)

4=a(3)(-4)

\dfrac{4}{(3)(-4)}=a

-\dfrac{1}{3}=a

Putting a=-\dfrac{1}{3}, we get

h(x)=-\dfrac{1}{3}x(x-7)              ...(ii)

The center of the parabola is the midpoint of 0 and 7, i.e., 3.

The width of the truck is 4 m. If is passes through the center then the truck must m 2 m on the left side of the center and 2 m on the right side of the center.

2 m on the left side of the center is x=1.5.

A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is possible if h(1.5) is greater than 2.8.

Putting x=1.5 in (ii), we get

h(1.5)=-\dfrac{1}{3}(1.5)(1.5-7)

h(1.5)=-(0.5)(-5.5)

h(1.5)=2.75

It is clear that h(1.5)<2.8, therefore the trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is not possible.

4 0
3 years ago
I have no idea!!! Help??
Savatey [412]

Answer:

x<5

Open circle at 5  going to the left

x >1

Open circle at 1 going to the right

Step-by-step explanation:

7x -19 < 16

Add 19 to each side

7x -19 +19< 16+19

7x< 35

Divide by 7

7x/7 < 35/7

x<5

Open circle at 5  going to the left

9+3x>12

Subtract 9 from each side

9-9 +3x >12-9

3x >3

3x/3 >3/3

x >1

Open circle at 1 going to the right

3 0
3 years ago
Question regarding logarithms.
Eddi Din [679]

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\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}

\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}

6 0
3 years ago
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