Draw a Rectangle with 20 units

Maureen spent $48 on books. That is 70% of her savings. What is the total amount of Maureen's savings? Round to the nearest penn

luda_lava [24]

The total amount of Maureen savings is $ 69

<em><u>Solution:</u></em>

Given that Maureen spent $ 48 on books. That is 70 % of her savings

To find: Total amount of Maureen savings

From given statement, 70 % of her savings is spent on books

Let "x" be the total amount of Maureen savings

Therefore,

70 % of total amount = $ 48

70 % of x = 48

Here "of" represents multiplication or product

$70 \% \times x = 48\\\\\frac{70}{100} \times x = 48\\\\0.7x = 48\\\\x = \frac{48}{0.7}\\\\x = 68.57 \approx 69$

Thus total amount of Maureen savings is $ 69

6 0

4 years ago

Suppose that, for every lot of 100 computer chips a company produces, an average of 1.4 are defective. Another company buys many

Vesna [10]

Answer:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)$

And we can find the individual probabilities like this:

$P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466$

$P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452$

$P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417$

$P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128$

And replacing we got:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda=1.4$ represent the average ocurrence rate per unit of time.

Solution to the problem

For this case the batch would be rejected if we found more than 3 defects, so then the probability of accept the batch would be given by:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)$

And we can find the individual probabilities like this:

$P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466$

$P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452$

$P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417$

$P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128$

And replacing we got:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463$

5 0

3 years ago

The height of a tree in feet over x years is modeled by the function f(x).

SVEN [57.7K]

The following statements are true:
The tree grows approximately 7 feet between years 5 and 7;
the tree stops growing at around 30 feet

In graphing the function, we see that the initial value, when x = 0, is 1 ft, not 2.
Additionally, we can see the values as the height increases; when the tree hits 15 feet, it continues to grow at around 1.5 feet every 0.4 years until it hits around 25 feet tall. Then it begins to slow down.

6 0

4 years ago

Read 2 more answers

1. What are the first 5 multiples of 9

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9, 18, 27, 36, 45, 54, 63, etc....

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4 years ago

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What is the answer for -16/9 = -4/3(5/3+n)

serious [3.7K]

N = -1/3 is the answer to the problem

8 0

4 years ago