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Roman55 [17]
3 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]3 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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Which is the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively?
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Which is the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively? 204.

Step-by-step explanation:

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The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th
inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4   -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

y(y-2)=100

y^2-2y=100

subtracting 100 from both sides

y^2-2y-100=0

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

ax^2+bx+c=0

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

Solving first

y=\frac{2+\sqrt{4+400}{2}

y=\frac{2+\sqrt{404}{2}

y=\frac{2+20.099}{2}

y=\frac{22.099}{2}

y=11.049

Solving second one

y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

y=\frac{2-\sqrt{4+400}{2}

y=\frac{2-\sqrt{404}{2}

y=\frac{2-20.099}{2}

y=\frac{-18.99}{2}

y=-9.045

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

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