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2 years ago

Complete the statement to describe how to convert ounces to pounds?

Mathematics

1 answer:

Digiron [165]2 years ago

6 0

Answer:

to find the weight in pounds, divide the number of ounces by the unit rate, 16 ounces per pound.

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The manufacturer of official NFL footballs uses a machine to inflate its new balls to a pressure of 13.5 pounds. When the machin

Dmitriy789 [7]

Answer:

106

Step-by-step explanation:

Given that:

Mean (m) = 13.5 pounds

Error margin = 0.025

α = 99%

Variation interval = range = 13.3 to 13.7

Standard deviation (σ) estimate from range :

Range / 4

(13.7 - 13.3) /4

= 0.4 / 4

= 0.1

Zcritical at 99% = 2.576

Using the formula :

Sample size (n) = [ (Zcritical * σ) / M. E]^2

n = [(2.576 * 0.1) / 0.025]^2

n = (0.2576 / 0.025)

n = 10.304^2

n = 106.172

n = 107 samples

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2 years ago

31 billion in scientific notation

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Before she started hiking, Jessica’s altimeter read +250 meters. She climbed 200 meters and reached point A, which is at an elev

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3 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:

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

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

Hence, the amount of oil was decreasing at 69300 barrels, yearly

7 0

2 years ago