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NeX [460]
3 years ago
13

Solve for substitution x+2y=-1 2x+3y=0

Mathematics
1 answer:
kipiarov [429]3 years ago
7 0
Hope my document helped! :)

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What is the value of x?<br><br><br><br> Enter your answer in the box.<br><br> x =
FinnZ [79.3K]

Answer:

x = 12

Step-by-step explanation:

IMPORTANT: Both answers should add up to 180 if they dont you did it wrong!

(10x - 20) + (6x + 8 ) = 180 plug in 12 for both of them

(10 * 12 - 20) + 6 * 12 + 8) = 120

(120 - 20) + (72+ 8 ) = 180

(100) + (80) = 180

so x = 12

hope this  helps please make brainliest

4 0
3 years ago
16x-5y=-33 solve the equation
LUCKY_DIMON [66]
 16x - 5y = -33               First, you subtract 16x from both sides of the 
<u>-16x           -16x       </u>      equation
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3 0
3 years ago
A) SSS<br> B) SAS<br> C) ASA<br> D) AAS
amm1812
This is B.SAS (side angle side)
4 0
4 years ago
The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s
Andrew [12]

Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that \mu = 1262, \sigma = 118

a) Determine the 26th percentile for the number of chocolate chips in a bag. ​

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

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

-0.643 = \frac{X - 1262}{118}

X - 1262 = -0.643*118

X = 1186

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

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

-1.96 = \frac{X - 1262}{118}

X - 1262 = -1.96*118

X = 1031

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

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

1.96 = \frac{X - 1262}{118}

X - 1262 = 1.96*118

X = 1493

Between 1031 and 1493.

​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

-0.675 = \frac{X - 1262}{118}

X - 1262 = -0.675*118

X = 1182

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

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

0.675 = \frac{X - 1262}{118}

X - 1262 = 0.675*118

X = 1342

IQR:

1342 - 1182 = 160

7 0
3 years ago
Which of the following is not a diameter?<br><br>SV<br>RU <br>WT
snow_tiger [21]
WT
It does not perfectly cut the circle in half.
7 0
3 years ago
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