How to change 611/17 to improper fractions

Each pound of fruit costs $4. Write an expression that shows the total cost of the fruit. Use the variable you identified in que

vovangra [49]

Answer:

$ $4f$

Step-by-step explanation:

Let's assume $f$ represents the number of pounds of fruit.

We need to multiply the number of pounds of fruit by the cost per pound.

This is $ $f * 4$ , or in a simpler form, $ $4f$ .

6 0

3 years ago

Which of the following steps would you perform to the system of equations below so that the equations have opposite y-coefficien

denis23 [38]

I think B not 100%shure ,you could prob go ask this on yahoo and get a bit more help ;) sorry I couldent help more

3 0

3 years ago

Read 2 more answers

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

3 0

3 years ago

What is the end behavior of the graph of

lutik1710 [3]

The end behavior of the graph of f(x) = -0.25x2 - 2x + 1 is that A) As x increases, f(x) increases. As x decreases, f(x) decreases.

<h3>What is the end behavior about?</h3>

The end behavior of a function f tells us that the way that the graph function behaves at the "ends" of the x-axis.

Note that the end behavior of a function tells the trend of the graph and in this graph, f(x) -----> +∞ as x----> -∞ (as the value of x decreases the value of f(x) increases) and f(x) -----> -∞ as x----> +∞ (as the value of x increases the value of f(x) decreases).

Therefore, The end behavior of the graph of f(x) = -0.25x2 - 2x + 1 is that A) As x increases, f(x) increases. As x decreases, f(x) decreases.

See full question below

What is the end behavior of the graph of f(x) = -0.25x2 - 2x + 1?

A) As x increases, f(x) increases. As x decreases, f(x) decreases.

B) As x increases, f(x) decreases. As x decreases, f(x) decreases.

C) As x increases, f(x) increases. As x decreases, f(x) increases.

D) As x increases, f(x) decreases. As x decreases, f(x) increases

Learn more about end behavior from

brainly.com/question/1365136

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