X increased by 61% plz help

Is it greater than or less than when the bubble on the graph line isn't colored in?

Andreas93 [3]

The bubble on the graph doesn't determine whether it is greater than or less than. when the bubble on the graph line isn't coloured in it means it is greater than or less than. when it is coloured in means it is greater than or equal to or less than or equal to.

8 0

3 years ago

What is f(x) = –4x2 + 24x + 13 written in vertex form? f(x) = –4(x – 6)2 + 23 f(x) = –4(x – 6)2 + 7 f(x) = –4(x – 3)2 + 4 f(x) =

nika2105 [10]

We have that
f(x) = –4x² + 24x + 13

we know that

The vertex form for a parabola that opens up or down is:

f(x) = a(x - h)^2 + k

in the given equation, a=-4, therefore we add zero to the original equation in the form of 4h²−4h²
f(x) = –4x² + 24x + 4h²−4h² +13
Factor 4 out of the first 3 terms and group them
f(x) = –4*(x² -6x +h²) +4h² +13
We can find the value of h by setting the middle term equal to -2hx
−2hx=−6x
h=3 and 4h²=36
f(x) = –4*(x² -6x +9) +36 +13

we know that the term (x² -6x +9) is equals to------> (x-3)²
so
f(x) = –4*(x-3)² +49

the answer is
f(x) = –4*(x-3)² +49

4 0

4 years ago

A researcher wants to know if the average time in jail for robbery has increased from what it was when the average sentence was

kati45 [8]

At 95% confidence interval, the z score is 1.96. Using the formula:

Confidence Interval = x ± z s / sqrt (n)

where x is the average time served, s is standard deviation and n is number of samples

Confidence interval = 7.5 ± (1.96 * 3) / sqrt (400)

Confidence interval = 7.5 ± 0.294

Confidence interval = 7.206 to 7.794

5 0

4 years ago

Please urgent! will give brainliest! Solve for x in simplest form. 7=1/2(x+2)

4vir4ik [10]

x=12 hopefully it helps

6 0

3 years ago

If $450 is invested at 6% compounded A (annually), B (quarterly), C (monthly), what is the amount after 7 years? How much intere

ss7ja [257]

Answer:

Step-by-step explanation:

Here's the gameplan for this. First of all we need a general formula, then we will define the variables for each.

The general formula for all of these is the same:

$A(t)=P(1+\frac{r}{n})^{nt}$

where A(t) is the amount after the compounding, P is the initial investment, n is the number of compoundings per year, r is the interest rate in decimal form, and t is time in years.

Then after we find the amount after the compounding, we will subtract the initial amount from that, because the amount at the end of the compounding is greater than the initial amount. It's greater because it represents the initial amount PLUS the interest earned. The difference between the initial amount and the amount at the end is the interest earned.

For A:

A(t) = ?

P = 450

n = 1

r = .06

t = 7

$A(t)=450(1+\frac{.06}{1})^{(1)(7)}$

Simplifying gives us

$A(t)=450(1.06)^7$

Raise 1.06 to the 7th power and then multiply in the 450 to get that

A(t) = 676.63 and

I = 676.63 - 450

I = 226.63

For B:

A(t) = ?

P = 450

n = 4 (there are 4 quarters in a year)

r = .06

t = 7

$A(t)=450(1+\frac{.06}{4})^{(4)(7)}$

Simplifying inside the parenthesis and multiplying the exponents together gives us

$A(t)=450(1.015)^{28}$

Raising 1.015 to the 28th power and then multiplying in the 450 gives us that

A(t) = 682.45

I = 682.45 - 450

I = 232.75

For C:

A(t) = ?

P = 450

n = 12 (there are 12 months in a year)

r = .06

t = 7

$A(t)=450(1+\frac{.06}{12})^{(12)(7)}$

Simplifying the parenthesis and the exponents:

$A(t)=450(1+.005)^{84}$

Adding inside the parenthesis and raising to the 84th power and multiplying in 450 gives you that

A(t) = 684.17

I = 684.17 - 450

I = 234.17

4 0

3 years ago