About how heavy is an adult camel?

Translate the expression One less than twice a number

Alja [10]

Answer:

2x – 1 = 17

Explanation:

Hope it helps you..

Y-your welcome in advance..

(;ŏ﹏ŏ)(ㆁωㆁ)

3 0

2 years ago

One side of a triangle is 2 times the second side. The third side is 5 ft longer than the second side. The perimeter of a triang

TiliK225 [7]

Answer:

The length of each side of triangle are 19 ft, 24 ft and 38 ft.

Step-by-step explanation:

Let the length of second side be x.

Now given:

Length of first side is 2 times length of second side.

Length of first side = 2x

Also, Length of third side is 5 ft longer than the second side.

Length of third side = 5+x

Perimeter of triangle = 81 ft.

We need to find the length of each side.

Now perimeter of triangle is sum of all three sides of triangle.

Therefore;

Perimeter of triangle = Length of first side + length of second side + Length of third side.

$2x+x+5+x=81 ft\\4x+5=81ft\\4x= 81 -5 ft\\4x = 76ft\\x= \frac{76}{4} = 19 ft$

Length of Second Side = 19 ft.

Length of First side = $2x= 2\times19 = 38 ft$

Length of Third side = $5+x= 5+19=24ft$

Hence the Length of triangles are 19 ft,38 ft,24 ft.

3 0

3 years ago

1) Find the missing side.

kari74 [83]

Answer:

25inch

Step-by-step explanation:

Ok so there is something called a Pythagorean theorem.

It's a²+b²=c²

It works on right triangles

C is the longest side or the side not touching the right angle

and a and b is the other 2 sides

so the missing side is 24²+7²=Missing side or c²

24²=576 and 7²= 49

576+49=625

$\sqrt{625}$ =25

3 0

3 years ago

How to subtract radians?

Dafna1 [17]

You first need to make sure the denominators are equal, then when they are, you can add/subtract the numerators. Then you can place the result over the common denominator.

I hope this helps! :)

4 0

3 years ago

Read 2 more answers

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