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3 years ago

Help! I don't understand this! Please respond ASAP

Mathematics

1 answer:

Umnica [9.8K]3 years ago

5 0

Answer:

To find a percent of the number, you need to convert the percentage into fractions or decimals. Then, multiply the fraction by the number.

Step-by-step explanation:

1. 60, 90,120,150,180,210,240,270

2. 65/100 x 200/1 = 2 x 65= 130\

3.5/100 x 180/1 = 9

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I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

professor190 [17]

Answer:

Step-by-step explanation:

The formula for the Pythagorean Theorem is:

$\sqrt{a^2+b^2}=c$

Now,

A=7

B=24

7x7=49

24x24=576

Add,

576+49=625

Last,

Take the square root of 625

The square root of 625 is 25

c=25

To Check if the hypotenuse is correct:

$a^2+b^2=c^2$

$7^2+24^2=25^2$

7x7=49

24x24=576

25x25=625

49+576=625

625=625,

Therefore the hypotenuse is 25.

5 0

2 years ago

Evaluate each function for the given value

Mariulka [41]

Answer:

3) 5

5) 1

Step-by-step explanation:

substitute the value of 9 for each 'x' in the expression

3) h(9) = 2/3(9) - 1

= 18/3 - 1

= 6-1 or 5

substitute the value of 5 for each 'x' in the expression

5) f(5) = -5² + 6(5) - 4

= -25 + 30 - 4

= 5-4 or 1

7 0

2 years ago

I have $45. I want to go to the store and buy some cheez-its. They are $2.75 a box. How many boxes can I get? Do I have any mone

nignag [31]

Answer:

You can by 16 boxes of cheez-its. You have 1 dollar left over.

Step-by-step explanation:

Divide 45 by 2.75 and you get 16.36, then multiply 2.75 by 16 to get 44.

So, you can pay 44 dollars for 16 boxes of cheez-its.

Hope this helps!!

Please mark Brainliest!

7 0

3 years ago

Read 2 more answers

Each of 100 students in the Allen School can only take 1 CSE class each, between the four classes CSE 311, CSE 312, CSE 331, and

Naddika [18.5K]

Answer:

$P(a=31,b=39,c=7,d=23) = 0.000668$

Step-by-step explanation:

Sample space, n = 100

Let the number of students signed up for CSE 311 = a

Let the number of students signed up for CSE 312 = b

Let the number of students signed up for CSE 331 = c

Let the number of students signed up for CSE 332 = d

Probability of taking CSE 311, $P_a$ = 0.3

Probability of taking CSE 312, $P_b$ = 0.4

Probability of taking CSE 331, $P_c$ = 0.1

Probability of taking CSE 332, $P_d$ = 0.2

$P(a,b,c,d) = \frac{n!}{a! b! c! d!} p_a^{a} p_b^{b} p_c^{c} p_d^{d} \\P(a=31,b=39,c=7,d=23) = \frac{100!}{31! 39! 7! 23!} * 0.3^{31} * 0.4^{39} * 0.1^{7} 0.2^{23}\\P(a=31,b=39,c=7,d=23) = \frac{4.58*10^{111}}{2.13*10^{56}* 5040 }* (1.57*10^{-55})\\P(a=31,b=39,c=7,d=23) = 0.000668$

3 0

3 years ago

Help me this is so difficult

Strike441 [17]

Answer:

Step-by-step explanation:

2 10

0 8

4 12

7 0

2 years ago