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snow_lady [41]
2 years ago
12

The lengths of the sides of an equilateral triangle are 3(x + )cm, (5x +2y-1) cm and 5x +1 Cm Find the side length of the triang

le. ​
Mathematics
1 answer:
pochemuha2 years ago
5 0

Answer:

Correct option is A

Since it is an equilateral triangle, the lengths of the sides are equal

So, x+3 y=3 x+2 y−2=4 x+

2

1

​

y+1

x+3 y=3 x+2 y−2

y=2 x−2

2 x−y=2 ...(1)

Also, 4 x+

2

1

y+1=x+3 y

8 x+y+2=2 x+6 y

6 x+2=5 y

6 x−5 y=−2   ...(2)

Multiplying e q(1) by  5 we get,

10 x−5 y=10      ...(3)

Subtracting e q(2) from e q(3) we get,

4 x=12⇒x=3

y=2 x−2=4

The length of one side of equilateral triangle = x+3 y =3+3(4)=15 units

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Answer:

b

Step-by-step explanation:

9x7.8x1/2x12

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What is 3/100 times 3
weeeeeb [17]

Answer:

3/100 times 3 is 0.09 or 9/100

7 0
2 years ago
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Determine which of the following equations, when graphed, intersect at the point (4, 0).
dexar [7]

Answer:

x - y = 4

x + y = 4

Step-by-step explanation:

Substitute x=4 and y=0 in every equation. If both sides of the equal sign are equal, (LS=RS, left side equals right side) then the graph intersects at (4,0).

- x - y = 4

- (4) - 0 = 4

-4 = 4

LS≠RS

x - y = 4

4 - 0 = 4

4 = 4

LS=RS

2 x + y = 7

2(4) + 0 = 7

8 = 7

LS≠RS

2 x + y = -7

2(4) + 0 = -7

8 = -7

LS≠RS

x + y = 4

4 + 0 = 4

4 = 4

LS=RS

2 x - y = 7

2(4) - 0 = 7

8 - 0 = 7

8 = 7

LS≠RS

Only x - y = 4  and x + y = 4  intersect at (4, 0).

4 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
Evan had 120 baseball cards. He gave 18 of them to his friend and the rest to his brother. What percent of the baseball cards di
jeka94

Answer:

85%

Step-by-step explanation:

Evan had 120 baseball cards. He gave 18 of them to his friend and the rest to his brother.

The number of baseball cards he gave to his brother is :

120 - 18 = 102 baseball cards

The percent of the baseball cards he gave to his brother is calculated as:

(Number of baseball cards given to his brother/ Total number of baseball cards) × 100

= 102/120 × 100

= 85%

The percent of baseball cards Evans gave his brother is 85%

3 0
2 years ago
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