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3 years ago

I don’t understand what Is the answer since I’m not good with percentages

Mathematics

1 answer:

TiliK225 [7]3 years ago

4 0

Answer:

9 students

75 percent of 12 = 9

Search for calculator. Type, "75" then click the percent symbol, then type "12." It should give you the right answer.

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Construct a stem-and-leaf plot of the test scores 67 comma 72 comma 85 comma 75 comma 89 comma 89 comma 87 comma 90 comma 99 co

VMariaS [17]

Answer:

Option C. Stem Leaves 6 7 7 2 5 8 5 7 9 9 9 0 9 10 0

Option D.

Step-by-step explanation:

The data values are 67, 72, 85, 75, 89, 89, 87, 90, 99 and 100.

Arranging the data values in ascending order

67, 72, 75, 85, 87, 89, 89, 90, 99, 100

The stem and lead plot can be shown under and stem is denoted as "S" whereas leaves are denoted as "L".

S L

6 7

7 2 5

8 5 7 9 9

9 0 9

10 0

The longer row of stem indicates the higher frequencies and so the length of rows are similar to the heights of bars in histogram.

5 0

2 years ago

How do I solve this?

iris [78.8K]

Answer:

Step-by-step explanation:

From the picture attached,

Addition of the blocks in first row is 60

a + a + a + 12 = 60

3a + 12 = 60

3a = 60 - 12

3a = 48

a = 16

For second row,

(b + 5) + (b + 5) + (b + 5) + (b + 5) = 60

4(b + 5) = 60

b + 5 = 15

b = 10

For third row,

(a + b) + c = (b + 5) + (b + 5) + (b + 5)

a + b + c = 3(b + 5)

16 + 10 + c = 3(10 + 5) [Since, a = 16 and b = 10}

26 + c = 45

c = 45 - 26

c = 19

For fourth row,

3c + d = 60

3(19) + d = 60

57 + d = 60

d = 60 - 57

d = 3

8 0

2 years ago

At a certain high school, 15 students play stringed instruments and

Aleks [24]

Answer:

0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

Step-by-step explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N( $A \cup B$ )' = 5

To find:

The probability that a randomly selected students plays both = ?

Solution:

Total Number of students = N(A)+N(B)+N( $A \cup B$ )' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

$N(A\cap B) =0$

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0$

So, 0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

5 0

2 years ago

A quadratic function is defined by g(x) = (x + 4)2 +7.

Travka [436]

Step-by-step explanation:

The vertex is the point (-4, 7).

4 0

2 years ago

1) Suppose f(x) = x2 and g(x) = |x|. Then the composites (fog)(x) = |x|2 = x2 and (gof)(x) = |x2| = x2 are both differentiable a

Rufina [12.5K]

Answer:

This contradict of the chain rule.

Step-by-step explanation:

The given functions are

$f(x)=x^2$

$g(x)=|x|$

It is given that,

$(f\circ g)(x)=|x|^2=x^2$

$(g\circ f)(x)=|x^2|=x^2$

According to chin rule,

$(f\circ g)(c)=f(g(c))=f'(g(c)g'(c)$

It means, $(f\circ g)(c)$ is differentiable if f(g(c)) and g(c) is differentiable at x=c.

Here g(x) is not differentiable at x=0 but both compositions are differentiable, which is a contradiction of the chain rule

3 0

3 years ago