All categories

2 years ago

(4+6)/(2•1) simplified

Mathematics

2 answers:

iris [78.8K]2 years ago

7 0

Answer:

Step-by-step explanation:

trust

Marizza181 [45]2 years ago

3 0

Answer:

(4+6)/(2•1)

(4+6=10)/(2×1=2)

10÷2

You might be interested in

Which expressions can be used to find the volume of the pyramid? Select three choices

Rashid [163]

Answer:

options????????

Step-by-step explanation:

...?.

8 0

3 years ago

a fisherman weighs an ahi tuna (a very large fish) on a scale and gets a reading of 135 pounds. the reading on the scale may hav

mamaluj [8]

Answer:

Coin 1 = quarter

Coin 2 = dine

Coin 3 = penny

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

In the university library elevator there is a sign indicating a 16-person limit as well as a weight limit of 2750 pounds. Suppos

Alona [7]

Answer:

0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit

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.

If n variables are added, the mean is $n\mu$ and the standard deviation is $s = \sqrt{n}\sigma$

In this problem:

$n = 16, \mu = 160*16 = 2560, s = \sqrt{16}*27 = 108$

What is the probability that the random sample of 16 people in the elevator will exceed the weight limit?

This is 1 subtracted by the pvalue of Z when X = 2750. So

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

By the Central Limit Theorem

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

$Z = \frac{2750 - 2560}{108}$

$Z = 1.76$

$Z = 1.76$ has a pvalue of 0.961

1 - 0.961 = 0.039

0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit

7 0

3 years ago

What is the value of x?<br><br> Enter your answer in the box.

ehidna [41]

Answer:

Step-by-step explanation:

c^2=a^2+b^2

where c = hypotenuse

so, x^2 = 24^2+7^2

x^2 = 576 + 49

x^2 = 625

x = square root of 625

x = 25

3 0

1 year ago

Read 2 more answers

A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a

sineoko [7]

ANSWER

The scale factor is
$2.5$

EXPLANATION

The given triangle has vertices,

$A(0,0),B(0,4),\:and\:C(6, 0)$ .

The vertices of the image triangle is,

$A'(0,0),B'(0,10),\:and\:C'(15, 0)$ .

The scale factor is given by

$k = \frac{image \: length}{object \: length}$

So we can use any of the corresponding sides to determine the scale factor,

$k = \frac{|A'B'|}{ |AB|}$

$k = \frac{ |10 - 0| }{ |4 - 0|}$

$k = \frac{ |10| }{ |4 |} = \frac{10}{4} = 2.5$

Or

$k = \frac{|A'C'|}{ |AC|}$

$k = \frac{ |15 - 0| }{ |6 - 0|}$

$k = \frac{ |15| }{ |6|} = \frac{15}{6} = 2.5$

Or

$k=\frac{|B'C'|}{|BC|}$

$k = \frac{\sqrt{(15 - 0)^2+(0-10)^2 }}{\sqrt{(6 - 0)^2+(0-4)^2}}$

$k = \frac{ 5\sqrt{13}}{2\sqrt{13}} = \frac{5}{2} = 2.5$

The correct answer is C

3 0

3 years ago