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2 years ago

Can y'all help me with my math problem please

Mathematics

1 answer:

ololo11 [35]2 years ago

8 0

• 2 x 10^5 = $200,000
• $200,000 * 8 years = $1,600,000
• $1,600,000 = 1.6 x 10^6

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1/4y =6

kondaur [170]

Answer: Y=24

Step-by-step explanation:1/4 y=6

*4 *4

y= 24

Math really is a universal language, Buenos noches amigo.

5 0

3 years ago

Anyone get this?? I really appreciated if you’ed help me :)

notsponge [240]

Answer:

V =5292 in ^3

Step-by-step explanation:

To find the volume for a rectangular prism

V = l*w*h

V = 6 * 36 * 24.5

V =5292 in ^3

6 0

3 years ago

The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability

amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565$

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

$Z = \frac{X - \mu}{\sigma}$

By the Central limit theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{70.5 - 69}{1.97565}$

$Z = 0.76$

$Z = 0.76$ has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0

2 years ago

If Richard can paint their living room in 4 hours, and Vanessa can paint the same living room in 5 hours, then how long will it

Tju [1.3M]

Answer:

2 hrs 13 mins

Step-by-step explanation:

If Richard can paint the living room in 4 hrs, then in 1 hour he can paint 1/4 of the living .

If Vanessa can paint the living room in 5 hours, then in one hour, she can paint 1/5 of the living room.

In one hour Richard and Vanessa can paint 1/4 + 1/5 = 1/y

y is the number of hours it will take both Richard and Vanessa to paint the living room. 1/y is how much of the house they can paint in one hour.

1/4 + 1/5 = 1/y

(5 + 4)/20. ( find the LCM)

9/20 = 1/y

9y= 20

y = 20/9

y = 2.22hrs

y = 2 hrs 13 mins

4 0

3 years ago

Where to put the ( )’s 5+4+3 2=29

Salsk061 [2.6K]

Answer:

It all depends.

Step-by-step explanation:

It depends on what you are trying to do, and what you want to solve for. If you are trying to distribute the 3 into 2, you would place the parentheses around the 2. I hope I was able to help, but there is not much that I can do without more clear instructions!

8 0

2 years ago