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aliina [53]
3 years ago
10

Graph 3x + y = -2 .................................................

Mathematics
1 answer:
nikdorinn [45]3 years ago
7 0

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

A graphing calculator does this very nicely.

__

The intercepts can be found by setting the other variable to zero and dividing by the coefficient of the variable of interest. This tells you the x-intercept is -2/3, and the y-intercept is -2/1 = -2. The y-intercept is conveniently on a grid point. To find other grid points on the line, it is convenient to choose integer values for x, then solve for y.

  y = -3x -2

Positive values for y will require negative values of x.

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14kg 450gram - 20kg = 5kg 550gram
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Step-by-step explanation:

4 0
3 years ago
I need help pls due tomorrow pls I just need part B done plsss
lara [203]

Part A

1 day = 1/4 hours of practice

7 days = 7/4 hours of practice (multiply both sides by 7)

1 week = 7/4 hours of practice

1 week = (4+3)/4 hours of practice

1 week = (4+3)/4 hours of practice

1 week = (4/4)+(3/4) hours of practice

1 week = 1+(3/4) hours of practice

1 week = 1 & 3/4 hours of practice

side note: 1 & 3/4 = 1.75

=======================================

Part B

Take the result from part A, and multiply it with 60

So we'll have 60 times 1&3/4 on the left side on the first line, then 60*(1+3/4) on the right side of this same line.

The rest of the lines look like this

(60*1) + (60*3/4)

60 + 60*3/4

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60 + 45

105 minutes

8 0
3 years ago
In the function f(x) = 3x + 2, f(5) =
Rudiy27

Answer:

17

Step-by-step explanation:

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6 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
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