Find the equation of the line below.

Find the least common multiple of 10x4 and 6w3 .

jarptica [38.1K]

10x⁴ and 6w³

1) We see no the same letters in both expressions, so we will need to include all letters in the final answer. x⁴w³

2) number part
10 = 2*5
6=3*2
We need 2*3*5=30 (we need 2 only one times, because we have 2 only one time in each of the numbers).
3)The least common multiple
2*3*5x⁴w³ = 30x⁴w³
Answer is 30x⁴w³.

8 0

3 years ago

He function f(d) = –3d + 60 gives the remaining life of a flashlight battery, in hours, d days after the start of a hike.

OLga [1]

D
The battery has 60 hours of life.

7 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

A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is ce

Scilla [17]

Answer: Assuming the riders starts at the position (20, 0) on the x-axis, the exact position of the rider will be (20cos75, 20sin75) or about (5.18, 19.32).

The angle for 5pi/12 radians is 75 degrees. Therefore, to find the position we can use the sine and cosine of 75 to find the x and y value of the coordinate.

For the y-value, we can write and solve:
sin75 = x/20

For the x-value, we can write and solve:
cos75 = x/20

3 0

3 years ago

A teacher gave her class two exams; 60% of the class passed the second exam, but only 48% of the class passed both exams. What p

Dahasolnce [82]

This is an example of conditional probability because we are trying to find the probability of an event occurring GIVEN the occurrence of some other event. There is a formula for this (see image attached).
If we follow this formula, the numerator would be the probability of (A AND B) which in this case is "48% of the class passed BOTH exams." The denominator in the formula would be that "60% of the class passed ONLY THE SECOND exam."
Therefore, P(A and B) = 0.48, which is 48% expressed as a decimal and P(B)= 0.60, which is 60% expressed as a decimal. Then, you can figure out the answer by dividing.

7 0

3 years ago

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