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r-ruslan [8.4K]
3 years ago
15

ILL GIVE BRAINLIEST HELP PLEASE!!! I only need the last question answered the formula is y=1.56x+1.29

Mathematics
2 answers:
denis-greek [22]3 years ago
7 0

Answer:

i got 32 weeks??

Step-by-step explanation:

raketka [301]3 years ago
3 0

Answer:

Graph the line using the slope and y-intercept, or two points.

Slope:

1.56

y-intercept:

Step-by-step explanation:

y-intercept:

(

0

,

1.29

)

x

y

−

0.827

0

0

1.29

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Find the area for this problem
Mashcka [7]

Answer:

the answer is 27

Step-by-step explanation:

3 0
3 years ago
Which of the following equations matches the graph above?
Brilliant_brown [7]

Option C:

The equation which matches the graph above is y = 4x - 4.

Solution:

Take any two points on the graph of the line.

Let the points are (0, -4) and (1, 0).

Here, x_1=0, y_1=-4, x_2=1, y_2=0

Slope of the line:

$m=\frac{y_2-y_1}{x_2-x_1}

$m=\frac{0-(-4)}{1-0}

$m=\frac{4}{1}

m = 4

y-intercept of the line is the point where the line crosses at y-axis.

Here, the line crosses at y-axis is (0, -4).

y-intercept, (c) = -4

Equation of a line is

y = mx + c

y = 4x + (-4)

y = 4x - 4

The equation which matches the graph above is y = 4x - 4.

Option C is the correct answer.

3 0
2 years ago
At a track meet, Tim ran at a constant speed of 300 meters per minute. After 8 minutes, he still had 600 meters left to run. The
gregori [183]
Okay so 300 meters per minute and he ran 8 minutes..

300x8=2400

He still had 600 left to go...

2400+600=3,000

He had to run three times longer than Eric...

3000/3=1000

Does that make sense?
8 0
3 years ago
A train must climb a constant gradient of 5.5m for every 200m on track. Find the angle of incline
Eduardwww [97]
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4 0
3 years ago
Read 2 more answers
Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars
GuDViN [60]

Answer:

Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

Step-by-step explanation:

We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.

<em>Let X = incomes for the industry</em>

So, X ~ N(\mu=95,\sigma^{2}=5^{2})

Now, the z score probability distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean income of firms in the industry = 95 million dollars

            \sigma = standard deviation = 5 million dollars

So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)

    P(X < 100) = P( \frac{X-\mu}{\sigma} < \frac{100-95}{5} ) = P(Z < 1) = 0.8413   {using z table]

                                                     

Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

5 0
3 years ago
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