Help me please help ASAP please please help me ASAP

Mathematics

1 answer:

IceJOKER [234]3 years ago

8 0

answer:

39.9

Step-by-step explanation:

54= area of rectangle

28.26= area of the circle

divide circle area by 2

= 14.13

subtract half of circle from rectangle area

54- 14.13

= 39.9

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Can anyone explain this to me please

timurjin [86]

Hello there,

To start it off, 2 x 39= 78
So a is correct that would be 2 x 30+9 = 2x39= 78
So b is correct as well. 2 x 40-1= 2 x 39= 78
c is questionable 2 x 30 +9= 60 + 9 = 69
So d is correct 18 + 60 = 78
So e is correct 100-22 = 78

So 2 x 39 is not equal to (2 x 30) +9

Hope this helped!

5 0

3 years ago

In Seaton Park school, 60% of the boys play baseball and 24% of the boys play baseball and football. What percentage of those th

Dmitrij [34]

Answer:

$0.4\%$

Step-by-step explanation:

Given: 60% of the boys play baseball and 24% of the boys play baseball and football in Seaton Park School

To find: percentage of those that play baseball also play football

Solution:

Let B denotes boys who play baseball and F denotes boys who play football.

$P(B)=60\%$

$P(F\cap B)=24\%$

Percentage of those that play baseball also play football = $P\left ( F|B \right )=\frac{P(F\cap B)}{P(B)}=\frac{24}{60}=\frac{2}{5}=0.4\%$

7 0

4 years ago

Sam ran 63,756 feet in 70 minutes. What is Sam's rate in miles per hour? ( there are 5,280 feet in one mile.)

umka21 [38]

Simply convert first feets into miles
Given is 5280 feets=1 miles

63756 /5280=12.075 miles

70 minutes = 1.16666= 1.17 hrs

rate is 12.075 miles/1.17 hrs

7 0

3 years ago

Read 2 more answers

Based on the figure below, the ratio of x/y is equivalent to which trigonometric function?

Ede4ka [16]

Answer: OPTION D.

Step-by-step explanation:

For this exercise it is important to remember the following Trigonometric Functions:

$sin\alpha =\frac{opposite}{hypotenuse}\\\\cos\alpha =\frac{adjacent}{hypotenuse}\\\\tan\alpha =\frac{opposite}{adjacent}\\\\sec\alpha =\frac{hypotenuse}{adjacent}\\\\csc\alpha =\frac{hypotenuse}{opposite}\\\\cot\alpha =\frac{adjacent}{opposite}$