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elena55 [62]
3 years ago
6

Samuel invested $53,000 in an account paying an interest rate of 6.7% compounded continuously. Assuming no deposits or withdrawa

ls are made, how much money, to the nearest cent, would be in the account after 17 years?
Mathematics
1 answer:
Natalija [7]3 years ago
4 0

Answer:

60367

Step-by-step explanation:

I = p x r x t

I = 53000 x 6.7/100 x 17

I = 60,367

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1/2(2+a)=3a+4/3 solve it
Mumz [18]
1/2(2+a)=3a+4/3

first, you use the distributive property<span>

</span>then your problem changes to...

1 + 1/2a = 3a + 4/3

then you subtract 1/2a with 3a

1  = 2 1/2a +4/3
<span>
now you subtract 1 with 4/3
</span>
-1/3 = 2 1/2a

now you divide -1/3 with 2 1/2a 

-2/15 = a

A = -2/15 is your answer. Hope this helps :)


4 0
4 years ago
8. Write an equation for the graph in slope-
MrMuchimi

Answer:

The answer is D

Step-by-step explanation:

the y is +10

slope is -1/3

8 0
3 years ago
Find the value of x when x² is equal to 121. * 4 points
Vaselesa [24]
You would take the square root of x^2 and of 121 so x =11
5 0
3 years ago
The area is 24 what are the possible dimensions l×w
taurus [48]
Hey there I see you need help well that's why I am here for a square the one side could be 6 or for a rectangle the l vould be 6 and the W could be 4 or 8 and 3 or lastly 12 and 2 hope this helps
5 0
4 years ago
Read 2 more answers
An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had
Stolb23 [73]

Answer:

Q1 z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the that means have no difference

Q2  CI 95 %  =  (  0,056 ;  0,164 )

Step-by-step explanation:

Sample information for people under 18

n₁  =  500

x₁ =  180

p₁  =  180/ 500    p₁  =  0,36    then  q₁  =  1 -  p₁     q₁ =  0,64

Sample information for people over 18

n₂  =  600

x₂  =  150

p₂  =  150 / 600   p₂ =  0,25   then   q₂  =  1 - p₂   q₂ =  1 - 0,25   q₂ = 0,75

Hypothesis Test

Null hypothesis                        H₀              p₁  =  p₂

Alternative Hypothesis           Hₐ              p₁  ≠  p₂

The alternative hypothesis indicates that the test is a two-tail test.

We will use the approximation to normal distribution of the binomial distribution according to the sizes of both samples.

Testin at CI =  95 %    significance level is  α = 5 %   α  =  0,05  and

α/ 2  =  0,025   z (c) for that α  is from z-table:

z(c) = 1,96

To calculate   z(s)

z(s)  =  ( p₁  -   p₂ ) / EED

EED = √(p₁*q₁)n₁  +  (p₂*q₂)/n₂

EED = √( 0,36*0.64)/500  +  (0,25*0,75)/600

EED = √0,00046  +  0,0003125

EED = 0,028

( p₁  -  p₂  )  =  0,36  -  0,25  = 0,11

Then

z(s)  =  0,11 / 0,028

z(s) = 3,93

Comparing  z(s) and  z (c)    z(s) > z(c)

z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the idea of equals means

Q2  CI  95 %   =  (  p₁  -  p₂  ) ±  z(c) * EED

CI 95%  =  ( 0,11   ±  1,96 * 0,028 )

CI 95%  = (  0,11  ±  0,054 )

CI 95 %  =  (  0,056 ;  0,164 )

8 0
3 years ago
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