The options in the drop downs are: the margin of error will decrease / the margin of error will increase

Mathematics

1 answer:

shtirl [24]2 years ago

8 0

Answer:

mark me brainliest come to my account i got you 100 B i swear all i do is giveaway B but u gotta give it a thanks.

Step-by-step explanation:

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What is the constant of variation for the quadratic variation? x 2 3 4 5 6 y 32 72 128 200 288

cluponka [151]

We use different models for different types of variation. For example, linear variation is associated with the formula y=ax, or the more familiar y=mx+b (the equation of a straight line). Cubic variation: y=a*x^3. In the present case we're discussing quadratic variation; perhaps that will ring a bell with you, reminding you that y=ax^2+bx+c is the general quadratic function. Now in y our math problem, we're told that this is a case of quadratic variation. Use the model y=a*x^2. For example, we know that if x=2, y =32. Mind substituting those two values into y=a*x^2 and solving for y? Then you could re-write y=a*x^2 substituting this value for a. Then check thisd value by substituting x=3, y=72, and see whether the resulting equation is true or not. If it is, your a value is correct. But overall I got 16!

7 0

3 years ago

Help me please?!!!!!!!!!

Basile [38]

I think it’s the last one but I’m not sure

4 0

3 years ago

Read 2 more answers

Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i

podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

$A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},$

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.$