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Stels [109]
2 years ago
11

Ayuda porfaaa JHAKJAH es para hoy y no entiendo un klo

Mathematics
1 answer:
Bond [772]2 years ago
3 0
A. +27
B. +4
C.+21

A. -16
B. 23
C. 3
D. -70
E. -20
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Solve for b 2 in A = h(b 1+b 2), if A = 16, h = 4, and b 1 = 3. b 2 =
hammer [34]

Answer:

b2 = 1

Step-by-step explanation:

A = h(b1 + b2)

Given:

A = 16

h = 4

b1 = 3

b2 = x

Work:

A = h(b1 + b2)

16 = 4(3 + x)

16 = 12 + 4x

4x = 16 - 12

4x = 4

x = 1

4 0
3 years ago
at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t
baherus [9]

The probability that a student participates in both sports and drama is 0.14.

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

If A and B are any two events, then the probability of the joint event (A\cup B) is given by the following formula: P(A\cup B)=P(A)+P(B)-P(A\cap B)

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose S denotes that "a student participates in sports" and D denotes that "a student participates in drama".
So, we have P(S)=\frac{42}{100}=0.42, P(D)=\frac{25}{100}=0.25, P(S\cup D)=\frac{53}{100}=0.53.

We want to find the probability that a student participates in both sports and drama i.e., we want to find P(S\cap D).

By the above formula, we obtain:

P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}

Therefore, the probability that a student participates in both sports and drama is 0.14.

To learn more about probability, refer: brainly.com/question/24756209

#SPJ9

6 0
2 years ago
Read 2 more answers
Enter the equation in slope-intercept form. Then graph the line described by the equation. -4x - 3y = 9
Alexus [3.1K]

Answer:

Step-by-step explanation:

-3y = 4x +9

y= -3/4x -3

5 0
3 years ago
2 The area of the back yard is 72m. The length is 12m. What is
meriva
2. 6m
3. No, she has 2.8ft of fabric. She is 2.2 yards short... :)
8 0
3 years ago
75% of the temperatures are below what value? How do you know? 75% of the temperatures are above what value? How do you know? Wh
Savatey [412]

Answer:

Complete Question:

Pick a city in Maryland and determine the average high temperatures for each month. Record this in a table.

a. Create a box-and-whisker plot for the data.

b. What is the median temperature?

c. 75%of the temperatures are below what value?How  do you know? d. 75% of the temperatures are above what value? How do you know? e. What conclusions can you draw about the temperature in Maryland?

Step-by-step explanation:

TASK 3

The city picked is Baltimore, the year 2017.

The table is attached in picture #1.

a) The box-and-whisker plot is attached in picture #2.

b) The median is the central value of the distribution i.e:  

42.4, 45.3, 46.8 | 54.7, 57.6, 66.0 | 69.1,  76.5,  80.6 | 85.8, 87.8, 90.0 

where we marked with a tally the quartiles.

As we have 12 values,  the average between the two central values will be the median :

M = (66.0 + 69.1) / 2 = 67.1 °F

c) 75% of the temperatures are below 83.2 °F.

Indeed, this value represents the third quartile. The position of the third quartile can be found by the formula:

3/4 · (n + 1) = 3/4 · (12 + 1) = 3/4 · 13 = 9.75

Therefore, since we did not get in integer position, the third quartile will be the average between the numbers in position 9 and 10:

q₀.₇₅ = (80.6 + 85.8) / 2 = 83.2 °F

d) 75% of the temperatures are above 50.8 °F.

Indeed, this value represents the first quartile. Similarly to point c), the position can be found by the formula:

1/4 · (n + 1) = 1/4 · 13 = 3.25

And therefore:

q₀.₂₅ = (46.8 + 54.7) / 2 = 50.8 °F. 

e) Baltimore in 2017 had a median high temperature of 67.1 °F.

25% of the year the high temperatures were warmer than 83.2 °F, with no outlier above 90.0 °F which was the hottest temperature.

25% of the high temperatures were colder than 50.8 °F, with no outlier below 42.4 °F, which was the coldest high temperature.

6 0
3 years ago
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