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dimulka [17.4K]
3 years ago
13

A scuba diver descends 48 feet in 4 mintues. What is the diver's average change in position per mintue relative to where she sta

rted?
Mathematics
1 answer:
marissa [1.9K]3 years ago
4 0
12 is the answer. you divide 48 by 4 to get 12
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Evaluate the expression |- 178|
Art [367]
The lines mean that the number is going to become positive. |-178| will become positive meaning it will become= |178|
3 0
3 years ago
Read 2 more answers
Please Help!!
Genrish500 [490]

Given

a\sqrt{x+b}+c=d

we have

\sqrt{x+b}=\dfrac{d-c}{a}

Squaring both sides, we have

x+b=\dfrac{(d-c)^2}{a^2}

And finally

x=\dfrac{(d-c)^2}{a^2}-b

Note that, when we square both sides, we have to assume that

\dfrac{d-c}{a}>0

because we're assuming that this fraction equals a square root, which is positive.

So, if that fraction is positive you'll actually have roots: choose

a=1,\ b=0,\ c=2,\ d=6

and you'll have

\sqrt{x}+2=6 \iff \sqrt{x}=4 \iff x=16

Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose

a=1,\ b=0,\ c=10,\ d=4

and you'll have

\sqrt{x}+10=4 \iff \sqrt{x}=-6

Squaring both sides (and here's the mistake!!) you'd have

x=36

which is not a solution for the equation, if we plug it in we have

\sqrt{x}+10=4 \implies \sqrt{36}+10=4 \implies 6+10=4

Which is clearly false

7 0
3 years ago
How can 6.75 + (-10.25) be expressed as the sum of its integer and decimal part
attashe74 [19]
There you go please brainliest thanks!

4 0
3 years ago
Solve the inequality for x: -3(2x+5) < -45
balandron [24]

Answer:

x>5

Step-by-step explanation:

-3(2x+5)<-45

2x+5<-45/-3

2x+5<15

2x<15-5

2x<10

x<10/2

x<5

x>5

7 0
3 years ago
An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in t
Natali5045456 [20]

Answer:

a) Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.

b) Rachel, because her time had the lowest z-score.

Step-by-step explanation:

Z-score:

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.

In this question:

The lower the time, the better the runner is. So whoever's time has a lower z-score is a better runner. So initially, we find the z-score of each student's time.

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in the class, ran 1 mile in 8 minutes.

This means that for Rachel's time, we have that X = 8, \mu = 11, \sigma = 3. So

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

Z = \frac{8 - 11}{3}

Z = -1

A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes.

This means that for Kenji's time, we have that X = 8.5, \mu = 9, \sigma = 2. So

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

Z = \frac{8.5 - 9}{2}

Z = -0.25

A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes.

This means that for Nedda's time, we have that X = 8, \mu = 7, \sigma = 4. So

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

Z = \frac{8 - 7}{4}

Z = 0.25

(a) Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?

Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.

(b) Who is the fastest runner with respect to his or her class? Explain why

Rachel, because her time had the lowest z-score.

8 0
3 years ago
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