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Anettt [7]
2 years ago
11

SOMEBODY HELP ME PLS!

Mathematics
1 answer:
babunello [35]2 years ago
3 0

Answer:

To find the value of y from a given value of x, find the position of x on the x-axis, then trace vertically until you meet the line.  Once you meet the line, trace horizontally to the y-axis to find the corresponding value of y.

To find the value of x from a given value of y, find the position of y on the y-axis, then trace horizontally until you meet the line.  Once you meet the line, trace vertically to the x-axis to find the corresponding value of x.

From inspection of the graph,

when x = 1,  y = -1

when x = 0, y = 2

when x = 2, y = -4

Therefore,

\begin{array}{| c | c |}\cline{1-2} x & y\\\cline{1-2} 1 & -1 \\\cline{1-2} 0 & 2 \\\cline{1-2} 2 & -4 \\\cline{1-2}\end{array}

To find the equation of the line, find the slope:

\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{2-(-1)}{0-1}=-3

Then use one of the points and the found slope with the point-slope form of a linear equation: y-y_1=m(x-x_1)

\implies y-2=-3(x-0)

\implies y=-3x+2

So the equation of this line is: y=-3x+2

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The 12 department chairs at Cedar H.S. represent the teaching staff. The principal assigns each chair a number and uses a random
Snezhnost [94]

Answer:

is

has

Step-by-step explanation:

8 0
3 years ago
Please help!!! 100 points if correct!!! please actually answer
rosijanka [135]

Answer:

see below

Step-by-step explanation:

What is the theoretical probability that the family has two dogs or two cats? (1/2)

The choices are dd, dc, cd, cc

There are 4 choices = dd or cc/ total = 2/4 = 1/2

Let the heads of the coin be dogs and the tails of a coin be cats

Flip two coins and coin A is the first pet and coin B is the second pet

heads , heads = 10

heads, tails  14

tails heads  =13

tails tails =  13

          total 50

Experimental probability  2 dogs or 2 cats = ( hh, tt) /total = ( 10+13) /50 =                    23/50

If we had 3 pets  

what is the theoretical probability that they have three dogs or three cats?

ddd, ddc, dcd, dcc, ccc, ccd ,cdc, cdd

There are 8 options

 ddd or ccc/ total = 2/8 = 1/4

Let the heads of the coin be dogs and the tails of a coin be cats

Flip three coins and coin A is the first pet and coin B is the second pet Coin C be the third pet

4 0
3 years ago
What is 4\5 times 60
ArbitrLikvidat [17]
We meet again
so first divide 60 by 5
this gives 12
then multiply 12 by 4
giving a final answer of 48
3 0
3 years ago
Read 2 more answers
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be 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.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0
4 years ago
If the area of a triangular kite is 6 square feet and its base is 4 feet, find the height
KonstantinChe [14]
The height of the kite is 3 feet. Here’s
why;
We know that BH=2A (base times height is equal to 2 times the area) of any given triangle.
Lets plug in values;
BH=2A
4(H)=2(6)
Then simplify;
4(H)=12
Flip the equation to find the missing value;
12 divided by 4 is 3.
Check your work;
4(3)=6(2):
12=12
The height of the kite is 3 feet.
4 0
3 years ago
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