Round 14.5 to a whole number

Which of the following expressions are equivalent to (9 7/8 + 2 4/5)-1/2

nikklg [1K]

Answer:A and B. Step-by-step explanation: C) 9 7/8 - (- 2 4/5) - (- 1/2) = = 9 7/8 + 2 4/5 + 1/2 (F). D) - (- 9 7/8 + 2 4/5) - 1/2) =.

I hope this helped^.^

May i have brainliest? ^.^

Have a great day buddy! ~hailey

7 0

2 years ago

Read 2 more answers

The total surface area of a cone whose radius is 2 r and slant height 2l is

vampirchik [111]

Answer:

$A=4\pi r(l+r)$

Step-by-step explanation:

The total surface area of a cone is given by :

$A=\pi rl+\pi r^2\\\\=\pi r(l+r)$

Put r = 2r and l = 2l in the above formula

$A=\pi \times 2r(2l+2r)\\\\A=2\pi r(2l+2r)\\\\=2\pi r \times 2(l+r)\\\\=4\pi r(l+r)$

Hence, the total surface area of the cone is $4\pi r(l+r)$ .

7 0

2 years ago

Find the slope of a line parallel to the line given

hram777 [196]

The answer is C hope this helped

5 0

3 years ago

15 POINTS. What is the area of the circle segment/shaded area? Picture included please help. Please answer in a way that will fi

PilotLPTM [1.2K]

I hope this helps you

8 0

3 years ago

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

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

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

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

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago