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3 years ago

Does anyone know how to slove this X+7=-3 X-y=-3

Mathematics

1 answer:

Agata [3.3K]3 years ago

8 0

X + 7 = - 3
x = - 3 - 7
x = - 10

Now put the value of x in 2nd equation,
- 10 - y = -3
- y = - 3 + 10
- y = 7
y = - 7

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Write the equation of the line that passes through the points (-2,3) and (-2,5),

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3 years ago

What is the slope of the line that passes through the points (-4, -1) and (4, 5)?

andreyandreev [35.5K]

Answer:

3/4

Step-by-step explanation:

Since we have two points, we can use the slope formula

m = ( y2-y1)/(x2-x1)

= ( 5 - -1)/( 4 - -4)

= ( 5+1)/(4+4)

= 6/8

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2 years ago

I need to simplify this expression : 3.2 + (b + 5.7 )

Leni [432]

The simplified expression is b+8.9

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3 years ago

PLEASE HELP ME IM BEGGING YOU IT IS EASY Consider the two groups listed below. Which statement describes the sets? (2 points) ho

miskamm [114]

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So, in order to have a valid function, any two points on the function cannot have the same x value that lands on different y.

With this property mentioned, it's clear to see that
(house, address) is NOT a function; however, (address, house) is a function, since same x lands on multiple y values is a valid onto function. (a house that have multiple addresses)

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3 years ago

Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies ac

Shtirlitz [24]

Answer:

a) $\mu -3\sigma = 336-3*6=318$

$\mu+-3\sigma = 336+3*6=354$

b) For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%

And then the % above would be 100-84= 16%

Step-by-step explanation:

Assuming this question : "Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies according to a roughly normal distribution with mean 336 days and standard deviation 6 days. Use the 68-95-99.7 rule to answer the following questions. "

(a) Almost all (99.7%) horse pregnancies fall in what range of lengths?

First we need to remember the concept of empirical rule.

From this case we assume that $X\sim N(\mu = 336. \sigma =6)$ where X represent the random variable "length of horse pregnancies from conception to birth"

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

From the empirical rule we know that we have 99.7% of the data within 3 deviations from the mean so then we can find the limits for this case with this:

$\mu -3\sigma = 336-3*6=318$

$\mu+-3\sigma = 336+3*6=354$

(b) What percent of horse pregnancies are longer than 342 days?

For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%

And then the % above would be 100-84= 16%

7 0

3 years ago