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3 years ago

Helllllllllllllllllpppppppppppppppppp

Mathematics

2 answers:

german3 years ago

5 0

Okay ????? With what doe

PIT_PIT [208]3 years ago

4 0

Answer:

Help with what?

Step-by-step explanation:

You might be interested in

A street light is at the top of a 25 ft pole. A 4 ft tall girl walks along a straight path away from the pole with a speed of 6

Burka [1]

Answer:

Tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

Step-by-step explanation:

Given : In the figure attached, Length of girl EC = 4 ft

Length of street light AB = 25 ft

Girl is moving away from the light with a speed = 6 ft per sec.

To Find : Rate ( $\frac{dw}{dt}$ ) of the tip (D) of the girl's shadow (BD) moving away from th

light.

Solution : Let the distance of the girl from the street light is = x feet

Length of the shadow CD is = y feet

Therefore, $\frac{dx}{dt}=6$ feet per sec. [Given]

In the figure attached, ΔAFE and ΔADE are similar.

By the property of similar triangles,

$\frac{x}{21}=\frac{x+y}{25}$

25x = 21(x + y)

25x = 21x + 21y

25x - 21x = 21y

4x = 21y

y = $\frac{4x}{21}$

Now we take the derivative on both the sides,

$\frac{dy}{dt}=\frac{4}{21}\times \frac{dx}{dt}$

= $\frac{4}{21}\times 6$

= $\frac{8}{7}$

≈ 1.14 ft per sec.

Since w = x + y

Therefore, $\frac{dw}{dt}= \frac{dx}{dt}+\frac{dy}{dt}$

$\frac{dw}{dt}=6+1.14$

= 7.14 ft per sec.

Therefore, tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

3 0

3 years ago

What is the answer and how do you even do this please help

Degger [83]

Answer:

Where is the question or problem that you want answered?

Step-by-step explanation:

I won't be of much help if I do not know what is it that you are asking help for.

7 0

3 years ago

Using the simple random sample of weights of women from a data set, we obtain these sample statistics: nequals45 and x overbar

Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : $n=45$

$\overline{x}=148.79\text{ lb}$

$\sigma=31.37\text{ lb}$

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = $\mu=148.79\text{ lb}$

b) The confidence interval for the population mean is given by :-

$\mu\ \pm E$ , where E is the margin of error.

Formula for Margin of error :-

$z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha=1-0.90=0.1$

Critical value : $z_{\alpha/2}=z_{0.05}=\pm1.645$

Margin of error : $E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69$

Now, the 90% confidence interval for the population mean will be :-

$148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)$

Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0

3 years ago

Can a smart brainstormer help me

lukranit [14]

The answer is .1391
Hope this helps!
(If your confused, I just put this in a calculator)

3 0

3 years ago

Read 2 more answers

A 4.5 pound of apples costs $22.50. how many pounds for a dollar?

AnnyKZ [126]

Answer:

.2 lbs

Step-by-step explanation:

4.5/22.5

4 0

3 years ago

Read 2 more answers