Please help me . factor. X^2+8x-33

For a normal distribution, what percentage of values lies within two standard deviations of the mean?

Semmy [17]

The answer is indeed D. If you check the normal distribution curve you get that within 1 standard diviation its
34%+34%=68%
Within 2 it's 34+34+13.5+13.5= 95%
And within 3 it's 34+34+13.5+13.5+2.5+2.5=100%

7 0

4 years ago

Dana graphs a point on a coordinate plane. The value of the x- coordinate of her is 5 more than the value of the y coordinate of

Natali5045456 [20]

The answer is c that what

7 0

4 years ago

you invest $1,000 in each of two accounts. account a earns simple interest at a rate of2.42% over 4 years. account b earns sim

Setler79 [48]

Answer:

Account A = $ 96.80

Account B = $ 48.40

Step-by-step explanation:

Account A

Principal = $1,000

rate = 2.42% = 0.0242

time = 4 years

To find the interest we will use the formula :

I = PTR

I = 1000 x 4 x 0.0242

I = $96.80

Account B

P = 1,000

t = 24 months = 2 years

r = 2.42% = 0.0242

I = PTR

I = 1000 x 2 x 0.0242

I = $ 48. 40

difference in interest = Interest a - Interest b

difference = 96.80 - 48.40

difference = $ 48 . 40

The interest on Account A doubles the interest on Account B

6 0

3 years ago

Read 2 more answers

The half-life of a certain radioactive substance is 45 days. There are 6.2 grams present initially. On what day

kvasek [131]

Answer:

There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days

Step-by-step explanation:

The given parameters are;

The half life of the radioactive substance = 45 days

The mass of the substance initially present = 6.2 grams

The expression for evaluating the half life is given as follows;

$N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}$

Where;

N(t) = The amount of the substance left after a given time period = 1 gram

N₀ = The initial amount of the radioactive substance = 6.2 grams

$t_{1/2}$ = The half life of the radioactive substance = 45 days

Substituting the values gives;

$1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$\dfrac{1}{6.2} = \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}$

$ln\left (\dfrac{1}{6.2} \right ) = {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )$

$t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days$

The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days

Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.

3 0

3 years ago

Without a calculator, order the following expressions from least to greatest. 9, pie squared, 3 times pie??

Elden [556K]

We have 9, 9.86, 9.42
So in order from least to greatest it would be: 9, 3 times pi, pi squared

5 0

3 years ago