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tresset_1 [31]
4 years ago
6

In the diagram, like a is the perpendicular bisector of km. what is the length of km

Mathematics
2 answers:
nikklg [1K]4 years ago
5 0

Answer:

hi did u attach the diagram?


Step-by-step explanation:


love history [14]4 years ago
4 0

would help with a diagram

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Jo is thinking of a number. four less than 9 times the number is 176. find jo's number
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The altitude of an isosceles triangle bisects the base and measures 10ft. If the base of the triangle measures 16ft, calculate:
Anarel [89]

Answer:

Step-by-step explanation:

the formula for calxulating the area if the isosceles triangle is expressed as;

A = \frac{1}{2}bh

b is the base of the triangle

h is the altitude

Given

b = 16ft

h = 10ft

Substitute

A = 1/2 * 16 * 10

A = 8*10

A = 80ft²

Hence the area of the triangle is 80ft²

b) To get the measure of one side of the triangle, we will use the pythagoras theorem. Let s be the slant side

s² = h²+(b/2)²

s² = 10²+(16/2)²

s² = 10²+(8)²

s² =100+64

s² =164

s = √164

s = 12.8ft

Hence one side of the triangle is 12.8ft

3 0
3 years ago
Lynne invested 35,000 into an account earning 4% annual interest compounded quarterly she makes no other deposits into the accou
storchak [24]

Hello!

Lynne invested 35,000 into an account earning 4% annual interest compounded quarterly she makes no other deposits into the account and does not withdraw any money. What is the balance of Lynne's account in 5years

Data:

P = 35000

r = 4% = 0,04

n = 4

t = 5

P' = ?

I = ?  

We have the following compound interest formula

P' = P*(1+\dfrac{r}{n})^{nt}

P' = 35000*(1+\frac{0,04}{4})^{4*5}

P' = 35000*(1+0,01)^{20}

P' = 35000*(1,01)^{20}

P' = 35000*(1.22019003995...)

P' \approx 42,706.66

So the new principal P' after 5 years is approximately $42,706.66.  

Subtracting the original principal from this amount gives the amount of interest received:

P' - P = I

42,706.66 - 35000 = \boxed{\boxed{7,706.66}}\end{array}}\qquad\checkmark

________________________

I Hope this helps, greetings ... Dexteright02! =)

4 0
3 years ago
Enter the range of values for x:<br> 2x - 4<br> 10<br> 45<br> 60<br> 2
Yakvenalex [24]

Step-by-step explanation:

hope this will help you.

8 0
3 years ago
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A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of
lina2011 [118]

Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean     x = 31,5

Sample standard deviation   s₂  = 8,35

Sample size        n₂ = 11

Ti-84 Calculator

Sample mean     y = 46,2

Sample standard deviation   s₁  = 9,99

Sample size        n₁ = 12

t(s)  = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s)  = 12,9 / √(99,8/12) + (69,72/11)

t(s)  = 12,9 / √8,32 + 6,34

t(s)  = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis               H₀      y - x > 1,80

Altenative Hypothesis       Hₐ      y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c)   critical value for

t(c) α; n₁ +n₂-2            df = 12 +11 -2     df  = 21

If we choose  CI = 95 %    then  α = 5 %   α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

8 0
3 years ago
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