What is the equation of the line graphed below?

kodGreya [7K]

Answer:

c=3 did you know that acording to the number letter the letter c is eqal to 3 and when said as c=3 looks like your special no no zone and c======3 looks like my special nono zone hahahahahahahahaahahahhahahahahahahhaahahahahhaahahaahahah

Step-by-step explanation:

_/(-o__-o)

8 0

3 years ago

If 8(a + b) = 48, what is the value of a + b?

Cloud [144]

Answer:

$8(a + b) = 48 \\ \\ a + b = \frac{48}{8} \\ \\ a + b = 6$

I hope I helped you^_^

5 0

3 years ago

Suppose I claim that the average monthly income of all students at college is at least $2000. Express H0 and H1 using mathematic

pishuonlain [190]

Answer:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): $\mu \geq 2000$

Alternative hypothesis (H1): $\mu$

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 200 or no

Step-by-step explanation:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): $\mu \geq 2000$

Alternative hypothesis (H1): $\mu$

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 2000 or no

7 0

3 years ago

4 Om obtuse, scalene Obright, scalene Oc, right, isosceles Od, right, equilateral

harina [27]

Answer:

B

Step-by-step explanation:

B. right, scalene

3 0

3 years ago

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