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LekaFEV [45]
4 years ago
11

Whats another way to express 52+24

Mathematics
2 answers:
Y_Kistochka [10]4 years ago
7 0

Answer: (50 + 2) + 20 +4

Step-by-step explanation: you can split each number into two parts that equal that number, like 52 could be 31 +21


BabaBlast [244]4 years ago
4 0

Answer:

4(13+6)

This is a method taught by my teacher. Its easy.

(4 x 13) + (4 x 6)

(52) + (24)


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How many triangles can be drawn with side lengths 3, 5, and 9?
prohojiy [21]

Answer:

no triangle

Step-by-step explanation:

SOLUTION: No; . The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

7 0
3 years ago
Find the unit price for each trail mix. cost is $6.00 and weight 3/4Lb
iragen [17]
6/ 3/4
6(4/3)
= 8

The answer is 8. 
5 0
4 years ago
The cost of 5 gallons of ice cream has a variance of 49 with a mean of 37 dollars during the summer. What is the probability tha
VMariaS [17]

Answer:

0.0119 = 1.19% probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected

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(which is the square root of the variance) \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 = 37, \sigma = \sqrt{49} = 7, n = 77, s = \frac{7}{\sqrt{77}} = 0.7977

What is the probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 35.2.

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

By the Central Limit Theorem

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

Z = \frac{35.2 - 37}{0.7977}

Z = -2.26

Z = -2.26 has a pvalue of 0.0119

0.0119 = 1.19% probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected

3 0
3 years ago
Evaluate 6²-7²+8²-9²+..+998²-999²​
Llana [10]

Answer:

-499,485.

Step-by-step explanation:

We can transform this to an arithmetic series by  working it out in pairs:

6^2 - 7^2 = (6-7)(6+7) = -13

8^2 - 9^2 = (8-9)*8+9) = -17  

10^2 - 11^2 = -1 * 21 = -21 and so on

The common difference is -4.

The number of terms in this series  is (998 - 6) / 2 + 1

= 992/2 + 1 = 497.

Sum of n terms of an A.S:

= n/2 [2a1 + (n - 1)d  

Here a1 = -13, n = 497, d = -4:

Sum = (497/2)[-26 - 4(497-1)]

= 497/2 * -2010

= -499,485.

3 0
4 years ago
Convert 3.8 kneecap to mi/min
shtirl [24]

keeping in mind that there 60 seconds in 1 minute, and about 1.609 kilometers in 1 mile.


\bf \cfrac{3.8~\underline{km}}{\underline{sec}}\cdot \cfrac{mi}{1.609~\underline{km}}\cdot \cfrac{60~\underline{sec}}{min}\implies \cfrac{3.8\cdot 60~mi}{1.609~min}~~\approx~~141.7029~\frac{mi}{min}

6 0
3 years ago
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