Which of the following functions is graphed below?

Please answer this correctly without making mistakes

Artyom0805 [142]

Answer:

6 tbsp and 2 tsp

Step-by-step explanation:

Hope this helps!

5 0

3 years ago

What 2 numbers add to get 2 and multiply to 4

rusak2 [61]

Factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4

first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x= $\frac{ -b+/-\sqrt{b^{2}-4ac} }{2a}$ so

1x^2-2x+4=0
a=1
b=-2
c=4
x= $\frac{ -(-2)+/-\sqrt{(-2)^{2}-4(1)(4)} }{2(1)}$
x= $\frac{ 2+/-\sqrt{4-16} }{2}$
x= $\frac{ 2+/-\sqrt{-12} }{2}$
we have $\sqrt{-12}$ and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x= $\frac{ 2+/-2\sqrt{-3} }{2}$
i= $\sqrt{-1}$
x= $\frac{ 2+/-2i\sqrt{3} }{2}$
x= $1+/-i\sqrt{3}$
x= $1-i\sqrt{3}$ or x= $1+i\sqrt{3}$ (those are the 2 numbers)

6 0

4 years ago

How much will $850 amount to be in three years if it is invested at 8% interest compounded quarterly for 3 years? A. $886.47 B.

Dimas [21]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{compounded amount}\\
P=\textit{original amount deposited}\to &\$850\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, so four times}
\end{array}\to &4\\

t=years\to &3
\end{cases}$

7 0

3 years ago

Please answer correctly !!!!!!! Will mark brainliest answer !!!!!!!!!!!!

Vesnalui [34]

The answer is d

Domain is just all x values

5 0

3 years ago

Suppose parts (a) through (d) below provide results for a study on the role of calcium in reducing the symptoms of PMS. For each

frutty [35]

Answer:

Step-by-step explanation:

Hello!

To test if calcium reduces the symptoms of PMS two independent groups of individuals are compared, the first group, control, is treated with the placebo, and the second group is treated with calcium.

The parameter to be estimated is the difference between the mean symptom scores of the placebo and calcium groups, symbolically: μ₁ - μ₂

There is no information about the distribution of both populations X₁~? and X₂~? but since both samples are big enough, n₁= 228 and n₂= 212, you can apply the central limit theorem and approximate the sampling distribution to normal X[bar]₁≈N(μ₁;δ₁²/n) and X[bar]₂≈N(μ₂;δ₂²/n)

The formula for the CI is:

[(X[bar]₁-X[bar]₂) ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} }$ ]