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Roman55 [17]
4 years ago
12

The test statistic of z equals = 2.37 2.37 is obtained when testing the claim that p greater than > 0.1 0.1. a. Identify the

hypothesis test as being​ two-tailed, left-tailed, or​ right-tailed. b. Find the​ P-value. c. Using a significance level of alpha α equals = 0.10 0.10​, should we reject Upper H 0 H0 or should we fail to reject Upper H 0 H0​?
Mathematics
1 answer:
Serjik [45]4 years ago
4 0

Answer:

Right tailed, 0.0089, reject.

Step-by-step explanation:

Given that The test statistic of z equals = 2.37  is obtained when testing the claim that p greater than > 0.1

we see that whenever claim is > it is right tailed.  And whenever p <alpha we reject H0

a) Since claim is p>0.1 this is right tailed test.

b) For z =2.37 p value is 0.5-0.4911\\0.0089

c)Since p < 0.01 our lapha we reject null hypothesis

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When distance decreases, gravity increases.

This means they have an inversely proportional relationship.

Step-by-step explanation:

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3 years ago
The diagram represents a flower border that is 3 feet wide
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Answer:

93

Step-by-step explanation:

108-15=93

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3 years ago
George has sold $75 worth of sandwiches at his concession stand today. Each sandwich sells for $5. He hopes to sell at least $60
boyakko [2]
5x + 75 > = 600 (thats greater then or equal)
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3 years ago
A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0
3 years ago
NEED HELP NOW!! Will give 20 extra PTS to the best answer
Harlamova29_29 [7]
To find the value of y, you need to plug a value into the x variable, in this case -8.

     4x + 9y = -14     Substitute -8 for x
4 (-8) + 9y = -14     Multiply
   -32 + 9y = -14     Start isolating the variable by adding 32 to both sides
            9y = 18      Divide both sides by 9
              y = 2

When x = -8, y = 2
4 0
3 years ago
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