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horsena [70]
3 years ago
13

2596 divided by 83 can't figure is out who ever does it gets $500 I really need help

Mathematics
1 answer:
worty [1.4K]3 years ago
8 0

Answer:

2596\83 =31.2

Step-by-step explanation:

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Which is the x-intercept of the graph of the equation y=3x−9?
padilas [110]
Y=3x−9

y=3(0)-9

y=9


Now plug in y to solve for x.

y=3x−9

9=3x-9

18=3x

6=x

The x-intercept is (6,0)

3 0
2 years ago
Angela practices piano for 5 hours on Sundays amd for 3 hours on all other days of the week. How many hours doea Angela practice
Kitty [74]

Answer:8 x 365=2920

Step-by-step explanation:

First, you add 5 + 3 which equals 8. Then there is 365 days in a year so you would multiply 8 times 365 which equals 2920.

Pretty sure this is right, hope it helps.

7 0
2 years ago
Solve : <br> 20-9x=11<br> 4(2x+1)=8<br> 5(2x+5)=-15<br> 3(8x+1)=-21
laila [671]
<span>20-9x=11
-9x=11-20
-9x= -9
x= -9/(-9)
x=1


4(2x+1)=8
2x+1=8/4
2x+1=2
2x=2-1
2x=1
x=1/2
x=0.5


5(2x+5)=-15
2x+5= -15/5
2x+5= -3
2x= -3-5
2x= -8
x= -8/2
x=-4

3(8x+1)=-21
8x+1= -21/3
8x+1= -7
8x= -7-1
8x =-8
x= -8/8
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</span>
4 0
3 years ago
2) A fair coin was tossed 75 times. The coin landed on Heads 42 times and Tails 33 times. What is the experimental probability o
coldgirl [10]

Answer:

\frac{11}{25}

Step-by-step explanation:

In the experiment that was conducted the coin was tossed a total of 75 times and out of those times it only landed on tails 33 times. Therefore the experimental probability of the coin landing on tails can be calculated by the dividing the times it landed on tails by the total number of times it was tossed. Like so...

\frac{33}{75} or 0.44

This fraction can also be simplified to its simplest form of \frac{11}{25} which is obtained by dividing both the numerator and denominator by 3

5 0
2 years ago
A given population proportion is .25. What is the probability of getting each of the following sample proportions
anyanavicka [17]

This question is incomplete, the complete question is;

A given population proportion is .25. What is the probability of getting each of the following sample proportions

a) n = 110 and = p^ ≤ 0.21, prob = ?

b) n = 33 and p^ > 0.24, prob = ?

Round all z values to 2 decimal places. Round all intermediate calculation and answers to 4 decimal places.)

Answer:

a) the probability of getting the sample proportion is 0.1660

b) the probability of getting the sample proportion is 0.5517

Step-by-step explanation:

Given the data in the questions

a)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 110

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 110 ) √( 0.1875 / 110 ) = 0.0413

Now, P( p^ ≤ 0.21 )

= P[ (( p^ - μ ) /S.D) < (( 0.21 - μ ) / S.D)

= P[ Z < ( 0.21 - 0.25 ) / 0.0413)

= P[ Z < -0.04 / 0.0413]

= P[ Z < -0.97 ]

from z-score table

P( X ≤ 0.21 ) = 0.1660

Therefore, the probability of getting the sample proportion is 0.1660

b)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 33

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 33 ) = √( 0.1875 / 33 ) = 0.0754

Now, P( p^ > 0.24 )  

= P[ (( p^ - μ ) /S.D) > (( 0.24 - μ ) / S.D)

= P[ Z > ( 0.24 - 0.25 ) / 0.0754 )

= P[ Z > -0.01 / 0.0754  ]

= P[ Z > -0.13 ]

= 1 - P[ Z < -0.13 ]

from z-score table

{P[ Z < -0.13 ] = 0.4483}

1 - 0.4483

P( p^ > 0.24 )  = 0.5517

Therefore, the probability of getting the sample proportion is 0.5517

6 0
2 years ago
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