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3 years ago

If 3 pounds of chocolate cost $25.50 how many pounds can you buy for $42.50

2 answers:

8 0

To solve this, write a proportion of Pounds to Cost:
Pounds: 3 x
Cost: 25.50 42.50

Cross multiply 3 and 42.50, this gives you 127.5
Then, divide by 25.5
This gives you 5.
You can buy 5 pounds with $42.50

irga5000 [103]3 years ago

5 0

Hnmm what's simple, follow this structure- always find the rate(by dividing) first find the rate of 3 pounds and 25.50
3pounds=. 1pound=?
25.50. Try for 1LBS So in order to get this divide 3 by the number that gets it 1 (for one pound)

Which is 3 so 3pounds divided by 3 = the one pound you want. So now divide the 25.50 by 3 to get 8.50. Now this is the problem
1pound=so ____pounds=
8.50. 42.50
All you have to do is divide 42.50 by 8.50 to get 5 FIVE POUNDS EQUAL 42.50 your welcome, I think :)

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Whats the area of the rectangle?

nevsk [136]

Answer:

Step-by-step explanation:

Area of rectangle = Length ×Breadth

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3 years ago

Need help due tomorrow please help

Mariulka [41]

Answer:

Step-by-step explanation:

It goes -1.33333, -2, -1, -.08, -.07 and that's incorrect.

3 0

3 years ago

Noah's oven thermometer gives a reading that is 2% greater than the actual temperature if the actual temperature is 350°F what w

blsea [12.9K]

Answer:

Step-by-step explanation:

Step one:

given data

actual reading= 350°F

thermometer reading 2% of the actual reading

Step two:

Required:

The thermometer reading

let us find 2% of 350°F

2/100*350°

=0.02*350

=7°F

Hence the thermometer reading is 7°F greater than 350°F

The thermometer reading is 350+7=357°F

6 0

3 years ago

A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and

RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

$A_t = A_0e^r^t$
where Aₜ = mass at time t, A₀ = mass at time 0, r = decay constant (rate), t = time

Our known variables are:

1998 to the year 2004 is a total of t = 6 years.
The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

$270=360e^-^r^(^6^)$
$270=360e^-^6^r$
$\frac{3}{4} =e^-^6^r$
$\text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)$
$\text{ln} (\frac{3}{4} )=-6r$
$r=-\frac{\text{ln}\frac{3}{4} }{6}$
$r=0.04794701$

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

$A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)$
$A_t=270e^-^0^.^3^3^5^6^2^9^0^7$
$A_t=193.01982213$

The expected mass of the sample in the year 2011 would be 193 mg.

3 0

3 years ago

The Community College Survey of Student Engagement reports that 46% of the students surveyed rarely or never use peer or other t

kap26 [50]

Answer:

d) No, this would not be unusual because 46% is only 1.2 standard errors from 40%.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If the z-score is higher than 2 or lower than -2, X is unusual.

In this question:

Mean = 40%. So $\mu = 0.4$