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expeople1 [14]
2 years ago
9

Lucy bought three books at a bookstore. Here are their prices (in dollars).

Mathematics
1 answer:
Valentin [98]2 years ago
8 0

Answer:

41.82$

Step-by-step explanation:

The prices are

7.58$, 16$, 18.24$.

To get the total of all what Lucy bought, You add the prices altogether.

That is,

18.24$

+16.00$

+<u>0</u><u>7</u><u>.</u><u>5</u><u>8</u><u>$</u>

41.82$

The total amount of what Lucy bought is 41.82$

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What is the value of x? round to the nearest hundredths place: <br> 0.45(x + 1.6) + 5x = 18
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Answer:

3.600

Step-by-step explanation:

0.25(x+1.6)+5x=18

Open the bracket first

0.25x+0.72+5x=18

Collect like terms

0.25x+5x=18-0.72

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Martin had 100 trees in his orchard the first year. Each year after that, he
alexgriva [62]

Answer:

177

Step-by-step explanation:

This scenario can be modeled as an <u>exponential function</u>.

General form of an exponential function: y=ab^x

where:

  • a is the initial value (y-intercept)
  • b is the base (growth/decay factor) in decimal form
  • x is the independent variable
  • y is the dependent variable

If b > 1 then it is an increasing function

If 0 < b < 1 then it is a decreasing function

If the number of trees <u>increase</u> by <u>10% each year</u>, then the number of trees each year will be 110% of the number of trees the previous year.  Therefore, the <u>growth factor is 110%</u>.

Given:

  • a = 100 trees
  • b = 110% = 1.10 (in decimal form)
  • x = time (in years)
  • y = number of trees in the orchard

Substituting the given values into the function:

\implies y=100(1.10)^x

(where x is time in years and y is the number of trees in the orchard)

To find how many trees are in the orchard in the 6th year, input x = 6 into the found equation:

\implies 100(1.10)^6=177.1561=177\: \sf (nearest\:whole\:number)

Therefore, Martin had 177 trees in his orchard in the sixth year.

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The incubation time for hummingbird eggs is approximately normal and has a mean of 16 days and standard deviation of 2 days. Let
Veseljchak [2.6K]

Answer:

0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 16 days and standard deviation of 2 days.

This means that \mu = 16, \sigma = 2

Probability that the length of hatching times is between 15 and 18 days.

This is the p-value of Z when X = 18 subtracted by the p-value of Z when X = 15. So

X = 18

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

Z = \frac{18 - 16}{2}

Z = 1

Z = 1 has a p-value of 0.8413

X = 15

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

Z = \frac{15 - 16}{2}

Z = -0.5

Z = -0.5 has a p-value of 0.3085

0.8413 - 0.3085 = 0.5328

0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.

4 0
3 years ago
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