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3 years ago

Help!! Please!!!!!

Mathematics

1 answer:

VLD [36.1K]3 years ago

4 0

Answer:

the mean is 8 and the absolute deviation is 1.5

Step-by-step explanation:

to find the mean you add up all the numbers, then you just divide by how many numbers there are. in this case, you would do 64/8 which would then equal 8. then you see how much the numbers differ from the mean. like for your numbers, for 8 it would be 0, for 9 it would be 1, for 7 it would be 1, for 6 it would be 2. you see it? its just how far away it is from the mean. so once you do that, you should have a 6. then you do 6/4 which is 1.5. hope this helps!

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Plz help for exam!!!<br> Solve for x and y.

ad-work [718]

Answer:

A x-198,y-99 $\sqrt{3}$

Step-by-step explanation:

since sin p/h you can say

perpendicular is 99

and hypotenuse is x

sin 30=99/x

x=198

and cos is b/h

base is y

and hypotenuse is x

cos 30=y/198

y-99 $\sqrt{3}$

6 0

3 years ago

Stephanie has a homeowners insurance policy for her $355,000 home with an annual premium of $0.42 per $100 of value and a deduct

frosja888 [35]

Answer:

$0.28 per 100 of value (B)

Step-by-step explanation:

Stephanie has a homeowner insurance policy of $355,000

Annual premium = $0.42 per 100

There is a deductible of $500

Stephanie has an annual out of pocket expense of

[($355,000/100) x $0.42] + $500 = $1,991

From the question, Stephanie now wants a new deductible amount of 1000.

Let X be the new annual premium

[(355,000X) / 100] + 1000 = 1991

3550X + 1000 = 1991

3550X = 1991 -1000

3550X = 991

X = 991/3550

X = 0.2791

X = 0.28 ( approximately)

The new annual premium is $0.28 per 100 of value

5 0

3 years ago

A store is having a sale on chocolate chips and walnuts. For 3 pounds of chocolate chips and 5 pounds of walnuts, the total cost

umka21 [38]

c = cost per pound of chocolate chips

w = cost per pound of walnuts.

$\bf \stackrel{\textit{3 lbs of "c"}}{3c}+\stackrel{\textit{5 lbs of "w"}}{5w}~~=~~\stackrel{\textit{costs}}{15} \\\\\\ \stackrel{\textit{12 lbs of "c"}}{12c}+\stackrel{\textit{2 lbs of "w"}}{2w}~~=~~\stackrel{\textit{costs}}{33} \end{cases}\qquad \impliedby \textit{let's use elimination} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llccccccl} 3c+5w=15&\times (-4)\implies &-12c&+&-20w&=&-60\\ 12c+2w=33&&12c&+&2w&=&33\\ \cline{3-7}\\ &&0&&-18w&=&-27 \end{array}$

$\bf -18w=-27\implies w=\cfrac{-27}{-18}\implies \blacktriangleright w=\cfrac{3}{2} \blacktriangleleft \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{3c+5\left(\cfrac{3}{2} \right)=15}\implies 3c+\cfrac{15}{2}=15 \implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( 3c+\cfrac{15}{2} \right)=2(15)} \\\\\\ 6c+15=30\implies 6c=15\implies c=\cfrac{15}{6}\implies \blacktriangleright c=\cfrac{5}{2} \blacktriangleleft$