H(0)=

A1,a2,a3....a30-each of these 30 sets has 5 elements.b1,b2,....bn-each of these n sets has 3 elements.union of a1,a2...a30=union

zzz [600]

the number of elements in the union of the A sets is:5(30)−rAwhere r is the number of repeats.Likewise the number of elements in the B sets is:3n−rB
Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have:150−9x=3n−8x⟺150−x=3n⟺50−x3=n
Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that:150 /10=15is the number of elements in the the A's without repeats counted (same for the Bs as well).So now we have:50−15 /3=n⟺n=45

5 0

3 years ago

A manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly.

denpristay [2]

Answer:

We conclude that the population mean light bulb life is at least 500 hours at the significance level of 0.01.

Step-by-step explanation:

We are given that a manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly. The population standard deviation is 50 hours and the light bulb life is normally distributed.

You select a sample of 100 light bulbs and find mean bulb life is 490 hours.

Let $\mu$ = population mean light bulb life.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 500 hours {means that the population mean light bulb life is at least 500 hours}

Alternate Hypothesis, $H_A$ : $\mu$ < 500 hours {means that the population mean light bulb life is below 500 hours}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean bulb life = 490 hours

σ = population standard deviation = 50 hours

n = sample of light bulbs = 100

So, the test statistics = $\frac{490-500}{\frac{50}{\sqrt{100} } }$

= -2

The value of z test statistics is -2.

Now, at 0.01 significance level the z table gives critical value of -2.33 for left-tailed test.

Since our test statistic is higher than the critical value of z as -2 > -2.33, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population mean light bulb life is at least 500 hours.

7 0

3 years ago

Which of the following are steps to use when formulating an equation? Check all that apply.

Ber [7]

Answer:

A and B

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

46 + (-2k) a. for k =3 b.for k = 23 k = -2

mamaluj [8]

9514 1404 393

Answer:

a. 40

b. 0

c. 50

Step-by-step explanation:

a) 46 + (-2k) for k = 3 is ...

46 + (-2(3)) = 46 -6 = 40

__

b) 46 + (-2k) for k = 23 is ...

46 + (-2(23)) = 46 -46 = 0

__

c) 46 + (-2k) for k = -2 is ...

46 + (-2(-2)) = 46 +4 = 50

3 0

2 years ago

the total cost of a belt and a jacket was $82.78 if the price of the belt was $0.86 less than the jacket what was the price of t

Aloiza [94]

B=bellt price
j=jacket price

b+j=82.78

belt is 0.86 less than jacket
b=j-0.86
add 0.86 to both sides
b+0.86=j
sub
b+b+0.86=82.78
2b+0.86=82.78
minus 0.86 from both sides
2b=81.92
divid both sides by 2
b=40.96

belt=$40.96

6 0

3 years ago

Read 2 more answers