All categories

3 years ago

a plant normally sells $6.50. today all plants are discounted by 25%. What is the sale price of the plant ?

Mathematics

2 answers:

Paraphin [41]3 years ago

5 0

Answer:

100% -> 6.50

100 - 25 = 75

75% -> 6.50/100 x 75

= 4.875

≈ 4.88

Cloud [144]3 years ago

3 0

Answer:

$4.875 or if rounded $4.88

Step-by-step explanation:

Hope this helped

also u just have to subtract 6.50 by 25%

You might be interested in

Bishaar's height is measured at 4.75 feet tall. Express Bishaar's height in meters. Round your answer to the nearest hundredth m

LuckyWell [14K]

To make the conversion, the first thing you should know is that
1m = 3.28ft
Then, doing the conversion we have
(4.75ft / 3.28ft) * (1m) = 1.448170732m
nearest hundredth
1.45m
anwser
Bishaar's height is1.45m

6 0

3 years ago

I'm a student at LUOA and, this is way too hard. Can someone help me, please?

puteri [66]

The answer is the first one.

Knowing that population density is like a rate, people per a given area, you would take the amount of people and divide by the area.

3.11E7/7.69E6 = 4.02 people per km squared.

Also the “E” is another way to write times 10 to a power and I recommend using that notation when plugging it into a calculator.

6 0

2 years ago

Help me with the correct answer

Kobotan [32]

Answer:

It’s A.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Condense the following logarithm: <img src="https://tex.z-dn.net/?f=5%20log_%7B4%7D%28a%29%20%20-%206%20log_%7B4%7D%28b%29%2

aleksley [76]

log 4 ( a^5 ) - log 4 ( b^6 ) =

log 4 ( a^5 / b^6 )

4 0

2 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago