All categories

3 years ago

Can someone help ill mark brainliest.

Mathematics

1 answer:

emmasim [6.3K]3 years ago

5 0

Answer:

y=3x+6

Step-by-step explanation:

When creating a slope equationg like this one the slope will always have the x next to it and the y intercept will always be on the end

You might be interested in

What is the slope of a line through (-5, -10) and (-1, 5)

Vinil7 [7]

Answer:

m = 15/4

Step-by-step explanation:

slope (m) = (y₂ - y₁)/(x₂ - x₁)

Let:

(x₁ , y₁) = (-5 , -10)

(x₂ , y₂) = (-1 , 5)

Plug in the corresponding numbers to the corresponding variables:

m = (5 - (-10))/(-1 - (-5))

m = (5 + 10)/(-1 + 5)

m = (15)/(4)

m = 15/4 is your answer.

3 0

3 years ago

How to solve 24.31÷2.6

Alecsey [184]

24.31 ÷ 2.6 = 243.1 ÷ 26 = 9.35

7 0

3 years ago

45.15 ÷ 21 what is the answer

Nastasia [14]

45.15 divided by 21 is 2.15. Hope it helps!!

3 0

3 years ago

Read 2 more answers

Please answer this ASAP

wlad13 [49]

f(x) = 5x + 1

f(4) = 5 * 4 + 1

f(4) = 20 + 1

f(4) = 21

g(x) = 2x² - 3x +1

g(5) = 5 * 5² - 3 * 5 + 1

g(5) = 125 - 15 + 1

g(5) = 111

f(4) + g(5) = 21 + 111 = 132

6 0

3 years ago

A book claims that more hockey players are born in January through March than in October through December. The following data sh

astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

3 0

4 years ago