The value of a boat is depreciating at a rate of 7% per year. In 2004 the boat was worth $180,000. Whats the value in 2009?

Meg is a veterinarian. In a given week, 50% of

krek1111 [17]

Answer:

a. Meg needs to find the part

b. Steve needs to find the percent.

Step-by-step explanation:

The question is incomplete.

This is the complete question:

Meg is a veterinarian. In a given week, 50% of the 16 dogs she saw were Boxers. Steve is also a veterinarian. In the same week, 7 of the 35 dogs he saw this week were Boxers. Each wants to record the part the whole, and the percent.

a. Does Meg need to find the part, the whole, or the percent.

b. Does Steve need to find the part, the whole, or the percent.

<h2>Solution to the problem</h2>

a. Does Meg need to find the part, the whole, or the percent.

Meg already knows the whole and the percent.

The whole is the entire number of dogs seen: 16 dogs.

The percent is 50%.

Hence, she needs to find the part: this is how many (what part) of the 16 dogs she saw were boxers.

And she can do with a simple operation:

Part = whole × percent = 16 × 50% = 8. This is 8 dogs of 16 were boxers.

b. Does Steve need to find the part, the whole, or the percent.

He already knows the part (7 of the 35 dogs were boxers) and the whole (35 dogs).

Then he needs to find the percent, which he can do using the next operation:

Percent = (part / whole) × 100 = (7/35)×100 = 20%

6 0

4 years ago

Helppppppppppppppppppppppp

Ratling [72]

9514 1404 393

Answer:

B, C

Step-by-step explanation:

The sums are ...

A) 59.45

B) 118.16

C) 18.97

D) 27.76

Sums B and C have 8 in the ones place.

__

You can add the ones-place digits. If the resulting ones-digit is 7 or 8, then add the tenths-place digits. If the carry from that sum, added to the sum of ones-place digits, gives a value of 8 or 18, then you found a sum with a ones digit of 8.

6 0

3 years ago

HELP ASAP!!!! PLEASE!!!! :(

aliya0001 [1]

120 degrees the sum of all is 720

8 0

3 years ago

Read 2 more answers

In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean

Nastasia [14]

Answer:

a) $\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

b) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n=80 represent the sample size

Solution to the problem

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

For this case we can calculate the mean like this:

$\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

Part b

For this case is the sample size is doubled the margin of error would be:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

5 0

3 years ago

Tell whether the rates are equivalent. 0.4 kilometer for every 15 minutes 0.6 kilometer for every 20 minutes

Ostrovityanka [42]

Answer:

no they are not equivalent

Step-by-step explanation:

0.4/15 = 0.026

so for every minute they travel 0.026km

0.6/20 = 0.03

so for every minute they travel 0.03km

so they are not the same/equivalent

6 0

3 years ago

The value of a boat is depreciating at a rate of 7% per year. In 2004 the boat was worth $180,000. Whats the value in 2009?​

The value of a boat is depreciating at a rate of 7% per year. In 2004 the boat was worth $180,000. Whats the value in 2009?