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Grace [21]
3 years ago
13

Write the equation of the line that is perpendicular to the line 5y=x−5 through the point (-1,0).

Mathematics
1 answer:
Artemon [7]3 years ago
7 0

Hey there!

First, we want to put the equation of this first line in slope-intercept form.

5y=x-5

We divide both sides by 5.

y=1/5x-1.

The slope of a perpendicular is line is the negative reciprocal of the slope of the original line. The slope of a line perpendicular to a line with the equation y=1/2x would be -2, because you flip the numerator and denominator and then make it negative.

So, this means that the slope of the line perpendicular to the line y=1/5x-1 is -5. So, here's our equation so far.

y= -5x+b

Now, we need to find the b. To do this, we can plug in this point (-1,0) that this perpendicular line goes through and solve for b.

0=-5(-1)+b

0=5+b

b= -5

So, this gives us the equation y= -5x-5

Have a wonderful day!

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Four more than three times a number is greater than twenty-two
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Answer:

3x + 4 > 22


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3 0
3 years ago
If x-2 is one of the factors of x^4-kx^2+4, find the value of k.
Andrei [34K]

Step-by-step explanation:

Given expression is x⁴-kx²+4 =0

x−2 is a factor of the given expression.

So, x=2

Substitute in the equation, we get

2⁴-k(2)²+4

⇒16+4k+4=0

⇒4k=−20

⇒k=−5

8 0
3 years ago
Juli buys a flat with 48 flowers. She was wants to plant them in equal rows of 5 flowers. How many rows can Juli plant? How many
miss Akunina [59]

Answer:

5 row, 3 left

Step-by-step explanation:

She has 48 flowers, 5 flowers one row

How many rows can she make? So divide

48/ 5

= 9.6

So she can only make 9 rows

The leftovers flowers are the flowers that didn't manage to make 5 per row

Therefore, if she can make 9 rows, in other words 9 x 5 = 45

45 flowers were able to be placed in a sequence of 5 per row

So the ones that are left is

48 - 45

= 3

8 0
3 years ago
Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa
xxMikexx [17]

Answer:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

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

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      32   18    50

Medium                                 68     7    75

Large                                     89    11    100

_____________________________________

Total                                     189    36   225

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

H0: independence between heath insurance coverage and size of the company

H1:  NO independence between heath insurance coverage and size of the company

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 col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*189}{225}=42

E_{2} =\frac{50*36}{225}=8

E_{3} =\frac{75*189}{225}=63

E_{4} =\frac{75*36}{225}=12

E_{5} =\frac{100*189}{225}=84

E_{6} =\frac{100*36}{225}=16

And the expected values are given by:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      42    8    50

Medium                                 63     12    75

Large                                     84    16    100

_____________________________________

Total                                     189    36   225

And now we can calculate the statistic:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

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

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0
3 years ago
Lauren and her family went to Maggiano's for dinner and spent $92.00. Lauren's family wanted to leave a 15% tip. How much will L
Andrews [41]

Answer:

105.8

Step-by-step explanation:

6 0
3 years ago
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