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2 years ago

What is the slope of the line below?

Mathematics

1 answer:

Flauer [41]2 years ago

6 0

Answer:

Undefined

Step-by-step explanation:

Vertical lines cannot be functions, since they yield multiple y-values for one single x-value.

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Create a table of data for two different linear functions. The table should use the same values of x for both functions. Based o

user100 [1]

Answer:

Yes they will intersect

Function 1= F(X)=2X+5

Function 2=H(X)=3X+2

INTERSECT=(3,11)

Step-by-step explanation:

First of all, we create 2 LINEAR function, i created the function f(x)=2x+5 and the function h(x)=3x+2, both are linear(without a quadratic term). Then

you replace the x for a number:

Table 1 (F(X)=2X+5) Table 2 (H(X)=3X+2)

X=1----->Y=2+5=7 X=1------>Y=3·1+2=5

X=2---->Y=2·2+5=9 X=2----->Y=3·2+2=8

X=3---->Y=3·3+5=11 X=3----->Y=3·3+2=11

With both tables of data we can see that in the X=3/Y=11 point this two linear functions will intersect so the answer is that the two functions will intersect at (3,11)----->(X,Y)

4 0

3 years ago

Round 541,776 to the nearest hundred thousand.

wel

500,000

The nearest hundred thousand would be 500,000 and 600,000, but since 541,776 has 4 in next unit, it rounds down.

6 0

3 years ago

Exterior angle theorem

Ivanshal [37]

$m1 = 79 + 58$

$m1 = 137$

7 0

1 year ago

Q={1,3,6,9,10,15} then rewrite in set builder method

AfilCa [17]

Answer:

{x:x is a multiple of 3, x ≤ 15}

4 0

1 year ago

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