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slega [8]
1 year ago
6

Does anybody know this? Find the slope of the line shown

Mathematics
1 answer:
alukav5142 [94]1 year ago
8 0

The slope of the linear function shown is of -1/3.

<h3>What is a linear function?</h3>

A linear function, in slope-intercept formula, is modeled by:

y = mx + b

In which:

  • m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

Given two points of a line, the slope is given by <u>change in y divided by change in x</u>.

In this problem, the line goes through points (-3,0) and (0,-1), hence the slope is given by:

m = (0 - (-1))/(-3 - 0) = 1/-3 = -1/3.

More can be learned about linear functions at

brainly.com/question/24808124

#SPJ1

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Two Answers: Angle 1, Angle 4

Adjacent angles share a common segment, line, or ray. Think of two adjacent rooms sharing a common wall.

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Option.B

Step-by-step explanation:

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Using this we can form an equation to find the value of x.

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3 years ago
Chris rents a car for a total of £375 The rental charges are £150 for the first day and £75 per day after that. For how many day
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Step-by-step explanation:

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3 years ago
Coldstone Creamery has a variety of different ice-cream flavors that they offer, and they have a variety of different serving si
Jobisdone [24]

Answer:

H0 : flavor and serving size are independent

H1: Flavor and serving size are not independent

Step-by-step explanation:

The claim or hypothesis is to test if there is a relationship between the different types of flavor (vanilla, strawberry and chocolate) and the size (large, medium and. Small) being ordered. This is the the Alternative hypothesis, in other words to establish that flavor type and size are not independent, (that is the two variables are correlated).

The Null hypothesis will be the opposite of the alternative, establish that both variables are independent.

Hence ;

Null hypothesis ; H0 : flavor and serving size are independent

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5 0
2 years ago
An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specifica
Natalija [7]

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = <u><em>diameter of the ball bearings</em></u>

SO, X ~ Normal(\mu=0.753,\sigma^{2} =0.004^{2})

The z-score probability distribution for normal distribution is given by;

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

where, \mu = population mean = 0.753 inch

           \sigma = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X \leq 0.75 inch)

      P(X < 0.753) = P( \frac{X-\mu}{\sigma} } } < \frac{0.753-0.753}{0.004} } } ) = P(Z < 0) = 0.50

      P(X \leq 0.75) = P( \frac{X-\mu}{\sigma} } } \leq \frac{0.75-0.753}{0.004} } } ) = P(Z \leq -0.75) = 1 - P(Z < 0.75)

                                                             = 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = <u>0.2734</u>.

(b) Probability that a ball bearing is between the  lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X \leq 0.74 inch)

      P(X < 0.75) = P( \frac{X-\mu}{\sigma} } } < \frac{0.75-0.753}{0.004} } } ) = P(Z < -0.75) = 1 - P(Z \leq 0.75)

                                                            = 1 - 0.7734 = 0.2266

      P(X \leq 0.74) = P( \frac{X-\mu}{\sigma} } } \leq \frac{0.74-0.753}{0.004} } } ) = P(Z \leq -3.25) = 1 - P(Z < 3.25)

                                                             = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = <u>0.226</u>.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

      P(X > 0.76) = P( \frac{X-\mu}{\sigma} } } > \frac{0.76-0.753}{0.004} } } ) = P(Z > -1.75) = 1 - P(Z \leq 1.75)

                                                            = 1 - 0.95994 = <u>0.0401</u>

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

      P(X < 0.74) = P( \frac{X-\mu}{\sigma} } } < \frac{0.74-0.753}{0.004} } } ) = P(Z < -3.25) = 1 - P(Z \leq 3.25)

                                                            = 1 - 0.9994 = <u>0.0006</u>

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

8 0
3 years ago
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