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svlad2 [7]
3 years ago
9

The executives of CareFree Insurance, Inc. feel that "a majority of our employees perceive a participatory management style at C

areFree." A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance. Eighty employees rate the management as participatory. The appropriate decision is __________.
Mathematics
1 answer:
Mandarinka [93]3 years ago
8 0

Answer:

Step-by-step explanation:

H_0: p = 0.5\\H_a : p >0.5

(One tailed test at 5% significance level)

Sample size =200 and sample proportion =\frac{80}{200} =0.40

Std error of p = \sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.4(0.6)}{200} } \\=0.0346

Test statistic = p diff/std error = -2.89

p value = 0.001926

So we reject H0 and accept alternate hypothesis.

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Mumz [18]

I'm not sure about my answer,but still writing it...

= ³√4a⁶ - 2/a-³ - (a²)3/2

= a² × ³√4 - 2a³ - 3a²/2

Cann't think that it could be futher more evaluted

8 0
3 years ago
Simplify.
elena55 [62]

\dfrac{4}{10}-\dfrac{2}{15}=\dfrac{4\cdot 15-10\cdot 2}{10\cdot 15}\\\\=\dfrac{40}{150}=\dfrac{4}{15}

The correct choice appears to be ...

... A) 4/15

7 0
3 years ago
Read 2 more answers
Find the value of x if 2x = 20 .​
7nadin3 [17]

Answer:

x = 10

Step-by-step explanation:

Divide both sides by 2.

(2x) / 2 = 20 / 2

x = 10

8 0
2 years ago
Read 2 more answers
Irunt
natita [175]

Answer:

30 = 2x + 3y

Step-by-step explanation:

x is the number of pens bought and y is the number of notbooks bought you graph this and you get a graph that shows you all the combinations.

8 0
3 years ago
Use induction to prove: For every integer n &gt; 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

2) Assume that f(k) is divisible by 5, f(k)=(k)^{5}-k

3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)      (divisible by 5)

add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

3 0
2 years ago
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