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tatuchka [14]
3 years ago
11

The graph of a system of equations with the same slope and the same y-intercepts will have no solutions.

Mathematics
2 answers:
Rasek [7]3 years ago
4 0
If 2 equations have the same y-intercept, they are overlapping, which means they have infinite solutions. So there is no way that 2 equations with the same y-intercept will have no solution. Thus your answer is: C)Never. 
Leokris [45]3 years ago
4 0
Never

if ur system has the same slope and the same y int, then there will be infinite solutions
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Josh is preparing for an exam by completing practice tests that have different numbers of questions. For each practice test, he
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The residuals are the numbers shown next to each black line. Just total them:


15 + 10 + 11-10 +9 +10 +3-17-12 = 19.


The answer is B. 19

8 0
3 years ago
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What is the missing constant term in the perfect square that starts with x2 – 16x?
andre [41]

Answer:

64

Step-by-step explanation:

the expression is

x² - 16x + _

the constant term is the square of the number residing with x divided by 2

the number number x = 16

therefore the constant term is (16/ 2)²

= 8²

= 64

the formula being used here is

(x + a)² = x² + 2ax + a²

4 0
3 years ago
Runner A completes 12 laps around a track in 22 minutes. Runner B completes 8 laps in 15 minutes. Which runner had a greater ave
castortr0y [4]

Answer:

2 divided by 22 equals 0.55. 8 divided by 15 equals 0.53. So runner A had a greater average speed.

Step-by-step explanation:

7 0
2 years ago
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Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t
Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let \sigma_{1}^{2} be the variance for the population of weight gains for rats given a low dose, and \sigma_{2}^{2} the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2} vs H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}. We have that the sample standard deviation for n_{2} = 22 female control rats was s_{2} = 28 g and for n_{1} = 18 female low-dose rats was s_{1} = 51 g. So, we have observed the value

F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176 which comes from a F distribution with n_{1} - 1 = 18 - 1 = 17 degrees of freedom (numerator) and n_{2} - 1 = 22 - 1 = 21 degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0
3 years ago
EXTRA POINTS---Set up an equation and find the value of x.
mars1129 [50]

Answer:

13) x = 30

14) x = 22.5

15) x = 36

Step-by-step explanation:

13) 4x+2x = 180

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x = 22.5

15) 2x+3x+2x+3x = 360

10x = 360

x = 36

8 0
3 years ago
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