All of the following values are part of the domain EXCEPT: -1 -3 -4

Two intelligent, honest students are sitting together at lunch one day when their math teacher hands them each a card. “your car

Lady bird [3.3K]

Answer: The loser's card shows 6.

Explanation: Let's start by naming the first student A and the second student B.
Since the product of A and B are either 12, 15, or 18, let's list every single possibility, the first number being A's number and the second number being B's number.

1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
4 3
5 3
6 2
6 3
9 2
12 1
15 1
18 1

Now, the information says that A doesn't know what B has, so we can immediately cross off all of the combinations that have the integer appearing once and once ONLY off, because if it happened once only, A would know of it straight away. Now, our sample space becomes much smaller.

1   12
1   15
1   18
2   6
2   9
3   4
3   5
3   6
6   2
6   3

Using this same logic, we know that we can cross off all of the digits that occur only once in B's column.

2   6
3   6

Now, A definitely knows what number B has because there is only one number left in B. Hence, we can conclude that the loser, B, has the integer 6.

4 0

3 years ago

if both expressions have the same value after substituting two different values and simplifying, then they are . When p = 2, the

Radda [10]

2p=16 and second expression is 8p=40

3 0

3 years ago

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GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

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 mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

Now, P-value of the test statistics is given by;

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

3 years ago

In ΔEFG, e = 210 cm, f = 520 cm and g=510 cm. Find the measure of ∠G to the nearest degree. PLEASE VERY URGENT.

Olegator [25]

Answer:

76°

Step-by-step explanation:

5 0

3 years ago

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36= 2² x 3² how do i get the number to my prime factor

mojhsa [17]

Answer:

add up the exponents on each prime number

Step-by-step explanation:

2^2= 2x2

3^2=3x3

so 36=2^2x3^2=2x2x3x3

5 0

3 years ago

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