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3 years ago

Find the equation of the line through the points (-3,-3) and (2,-1) using point-slope form. Then rewrite the

Mathematics

1 answer:

IrinaVladis [17]3 years ago

3 0

Answer: See below

Step-by-step explanation:

The point-slope equation is y-y₁=m(x-x₁). Since we don't know our slope, we can use the formula $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$ to find the slope. All we have to do is use the coordinate we were given and plug it into the formula.

$m=\frac{-1-(-3)}{2-(-3)} =\frac{2}{5}$

Now that we have the slope, we can fill out the point-slope equation.

y-(-3)=2/5(x-(-3))

y+6=2/5(x+3)

This is the point-slope form.

Now, we can distribute and solve to get slope-intercept form.

y+6=2/5x+6/5

y=2/5x-24/5

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Moneys taken out of a salary for such things as taxes, medical insurance, and retirement funds are called

Tamiku [17]

They are known as deductions, or deductibles,because they are deducted from your wage, before it goes into your bank account.
The answer is D
Hope this has helped you :)

4 0

3 years ago

Heather has divided $5600 between two investments, one paying 9% the other paying 5%. If the return investment is $396, how much

Westkost [7]

Answer:

x = 2700 y=2900

Step-by-step explanation:

x+y = 5600

5/100 x + 9/100 y = 396

x+y=5600

5x+9y=39600

5x+5y = 28000

5x+9y=39600

-4y = -11600

y= 2900

x = 5600-2900 = 2700

3 0

3 years ago

Read 2 more answers

What is a simpler form of the radical expression?

adoni [48]

∛125x^21y^24

Split them up:
∛125 * ∛x^21 * ∛y^24

∛125 = 5
∛x^21 = x^7 21/3 = 7
∛y^24 = y^8 24/3 = 8

The simpler form of the radical expression: ∛25x^21y^24 is: <span>D. 5x^7y^8</span>

5 0

4 years ago

A psychologist wants to see if a certain company has fair hiring practices in an industry where 60% of the workers are men and 4

makvit [3.9K]

Answer:

A) $H_{0}$ : p=0.5 (At least half of the workers are women,fair)

B) $H_{a}$ : p<0.5 (Less than half of the workers are women,unfair)

C) critical value of the test statistic is 1.64 (one tailed)

D) Test statistic is ≈ 0.29

E) Since 0.29<1.64, we fail to reject the null hypothesis. There is no significant evidence that the company has unfair hiring practices at 0.05 significance level.

Step-by-step explanation:

Let p be the proportion of women workers in the company. Null and alternative hypotheses are

$H_{0}$ : p=0.5 (At least half of the workers are women,fair)

$H_{a}$ : p<0.5 (Less than half of the workers are women,unfair)

Test statistic can be found using the equation:

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of women workers ( $\frac{55}{107} =0.514$ )

p is the proportion assumed under null hypothesis. (0.5)

N is the sample size (55+52=107)

Then z= $\frac{0.514-0.5}{\sqrt{\frac{0.5*0.5}{107} } }$ ≈ 0.29

For alpha 0.05, critical value of the test statistic is 1.64 (one tailed)

Since 0.29<1.64, we fail to reject the null hypothesis. There is no significant evidence that the company has unfair hiring practices at 0.05 significance level.

4 0

3 years ago

Let H be a subgroup of a group G. We call H characteristic in G if for any automorphism σ∈Aut(G) of G, we have σ(H)=H.

choli [55]

Answer:Problem 1. Let G be a group and let H, K be two subgroups of G. Dene the set HK = {hk : h ∈ H,k ∈ K}.

a) Prove that if both H and K are normal then H ∩ K is also a normal subgroup of G.

b) Prove that if H is normal then H ∩ K is a normal subgroup of K.

c) Prove that if H is normal then HK = KH and HK is a subgroup of G.

d) Prove that if both H and K are normal then HK is a normal subgroup of G.

e) What is HK when G = D16, H = {I,S}, K = {I,T2,T4,T6}? Can you give geometric description of HK?

Solution: a) We know that H ∩ K is a subgroup (Problem 3a) of homework 33). In order to prove that it is a normal subgroup let g ∈ G and h ∈ H ∩ K. Thus h ∈ H and h ∈ K. Since both H and K are normal, we have ghg−1 ∈ H and ghg−1 ∈ K. Consequently, ghg−1 ∈ H ∩ K, which proves that H ∩ K is a normal subgroup.

b) Suppose that H G. Let K ∈ k and h ∈ H ∩ K. Then khk−1 ∈ H (since H is normal in G) and khk−1 ∈ K (since both h and k are in K), so khk−1 ∈ H ∩ K. This proves that H ∩ K K.

c) Let x ∈ HK. Then x = hk for some h ∈ H and k ∈ K. Note that x = hk = k(k−1hk). Since k ∈ K and k−1hk ∈ H (here we use the assumption that H G), we see that x ∈ KH. This shows that HK ⊆ KH. To see the opposite inclusion, consider y ∈ KH, so y = kh for some h ∈ H and k ∈ K. Thus y = (khk−1)k ∈ HK, which proves that KH ⊆ HK and therefoere HK = KH. To prove that HK is a subgroup note that e = e · e ∈ HK. If a,b ∈ HK then a = hk and b = h1k1 for some h,h1 ∈ H and k,k1 ∈ K. Thus ab = hkh1k1. Since HK = KH and kh1 ∈ KH, we have kh1 = h2k2 for some k2 ∈ K, h2 ∈ H. Consequently,

ab = h(kh1)k1 = h(h2k2)k1 = (hh2)(k2k1) ∈ HK

(since hh2 ∈ H and k2k1 ∈ K). Thus HK is closed under multiplication. Finally,

Step-by-step explanation:

6 0

3 years ago