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3 years ago

What is the value of 5^3i^9?

1 answer:

3 0

That should be 125i. If you have a graphing calculator you should be able to just enter it into the calculator to solve lol

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Cody and Jordan were selling Flamin' Hot Cheetos to raise money for the Tornados. Cody sold 5 bags of Cheetos for every 3 bags J

andrew11 [14]

They sold 160 and for each 5 bags sold by one 3 bags are sold by another 3+5=8.

160/8=20.

Now Since Cody Sold 5 Bags for every 3 bags for Jordan

5x20=100

3x20=60

Cody Sold 100 and Jordan Sold 60

7 0

2 years ago

Read 2 more answers

Ridge wood bank is offering 5% simple interest on all of their saving accounts if an account it opened with a staring balance of

marta [7]

Answer:

$230

Step-by-step explanation:

plz mark brainliest

8 0

2 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago

Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and

Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437$

P(sample average sediment density is at most 3.00)

$P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)$

Calculation the value from standard normal z table, we have,

$P(x \leq 3.00) = 0.918$

0.918 is the probability that the sample average sediment density is at most 3.00

4 0

2 years ago

What digit has the thousandths value in two hundred forty-six ten thousandths? *

Fed [463]

The answer is D, 10 is in the thousandths place

8 0

2 years ago

Read 2 more answers