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3 years ago

The points A, B and C in 3D space are such that:

Mathematics

1 answer:

Korvikt [17]3 years ago

6 0

Answer:

Angle of elevation= 180- (20- 55)

180- 75

105°

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Expand each expression. ln (2x)4

8_murik_8 [283]

Answer:

4ln(2) + 4ln(x)

Step-by-step explanation:

ln(2x)⁴

4ln(2x)

4[ln(2) + ln(x)]

4ln(2) + 4ln(x)

4 0

3 years ago

Read 2 more answers

Solve for x. Round to the nearest tenth, if necessary.

Tcecarenko [31]

ANSWER

14.3

EXPLANATION

angle = 21⁰

opposite side = 5.5

adjacent side = x

Knowing this, we must use tangent (opposite / adjacent)

tan (21⁰) = 5.5 / x

x = 5.5 / tan (21⁰) (use calculator)

x= 14.3

4 0

2 years ago

Distance formula for (0,1) (3,7)

krok68 [10]

if those r coordinates you can use the Pythagorean the to find the distance

plot the point and create a right triangle with the origin

could up the spaces and depending on the number, subtract the side from the hypotenuse or add the sides to find the hypotenuse

then square root both sides

5 0

3 years ago

Select the correct answer.

Crazy boy [7]

Answer:

Step-by-step explanation:

to get from 3/4 to 12/x you gotta multiple 3 times 4 which is 12 so the bottom multiple that by 4 also which 4 x 4 is 16

3 0

3 years ago

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A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, 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 mean waiting time is less than 10 minutes.

5 0

3 years ago