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gayaneshka [121]
3 years ago
15

Write the slope-intercept form of the equation of each line given the slope and y-intercept.

Mathematics
1 answer:
mr Goodwill [35]3 years ago
7 0

Answer:

y = 3x+5 or x=-2 y=5

Step-by-step explanation:

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a company has decreased the weight of its boxes of macaroni by 8 %. if the new weight of the box is 15.1 ​ounces, what was the o
quester [9]
8%=0.08 and 100%=1 (100% represents the original weight in percentaje)
If x is the weight you can express: 1x-0.08x=15.1
solve for x:  0.92x=15.1
x= 15.1/0.92=16.1
The original weight was 16.41 ounces
8 0
3 years ago
Read 2 more answers
Suppose that the credit remaining on a phone card (in dollars) is a linear function of the total calling time (in minutes). When
Virty [35]

Answer:

Credit remaining after 21 minutes = $30.4

Step-by-step explanation:

Credit remaining on a phone card is a linear function of the total calling time.

When graphed, let the linear function representing the line is,

y = mx + b

Where 'm' = slope of the line

b = y-intercept

From the graph,

Slope of the line = -0.12

y = -0.12x + b

If this line passes through a point (33, 28.96),

28.96 = -0.12(33) + b

b = 28.96 + 3.96

b = 32.92

Therefore, the linear function is,

f(x) = -0.12x + 32.92

where x = calling time

Credit left in the card after 21 minutes,

f(21) = -0.12(21) + 32.92

      = -2.52 + 32.92

      = $30.4

6 0
3 years ago
I need help with this​
noname [10]

Answer:

with what

Step-by-step explanation:

5 0
3 years ago
Please help!! | show work and do not guess. :)
Olegator [25]
The answer would be b A≈1636.8
8 0
3 years ago
A researcher determines that students are active about 60 + 12 (M + SD) minutes per day. Assuming these data are normally distri
rjkz [21]

Answer:

The correct option is (b).

Step-by-step explanation:

If X \sim N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z \sim N (0, 1).

The distribution of these z-variate is known as the standard normal distribution.

The mean and standard deviation of the active minutes of students is:

<em>μ</em> = 60 minutes

<em>σ </em> = 12 minutes

Compute the <em>z</em>-score for the student being active 48 minutes as follows:

Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0

Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.

The correct option is (b).

4 0
3 years ago
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