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2 years ago

A profit of $9 in integers

Mathematics

2 answers:

DedPeter [7]2 years ago

4 0

Because the situation represents a profit, the integer is 9.

svetoff [14.1K]2 years ago

4 0

The integer is 9 because the situation represents a profit.

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Complete this pattern 3400000, 34000,______3.4,______

Ira Lisetskai [31]

Well the pattern is moving 2 decimal places to the left to make it smaller so it starts off with 3,400,000. 34,000. Then 340. 3.4 and then .034.
ANSWER: 1) 340 2) .034

3 0

3 years ago

Let f(x) = 1/x . Find the number b such that the average rate of change of f on the interval [2, b] is − 1/8

vekshin1

Answer:

b=4

Step-by-step explanation:

So, we have the function $f(x)=1/x$ . We need to find b such that the average rate of change or the slope is -1/8 between the intervel [2, b]. First, let's find f(2).

f(2) = 1/(2) = 1/2

So, we have the point (2, 1/2)

At point b, f(b) = 1/b.

Let's plug this into the slope formula:

$\frac{y_2-y_1}{x_2-x_1}=\frac{.5-\frac{1}{b} }{2-b} =-1/8$

Now, we just need to solve for b. First, let's multiply both the numerator and denominator by b (to get rid of the annoying fraction in the numerator).

$\frac{.5b-1}{2b-b^2} =\frac{-1}{8}$

Now, cross multiply.

$4b-8=b^2-2b$

$b^2-6b+8=0$

Solve for b. Factor using the numbers -4 and -2.

$=(b-4)(b-2)=0$

Thus, b=4 or b=2.

However, b=2 is not a possible solution since the interval [2,2] means nothing. Thus, b=4.

4 0

2 years ago

If there are 15 students who don’t wear glasses, how many students are in the class?

mixer [17]

Ummmm still 15???????

4 0

2 years ago

Tuto

alexgriva [62]

answer is 40%

hope it helped you

8 0

2 years ago

Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.

adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0

2 years ago