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nikitadnepr [17]
3 years ago
6

FIRST RIGHT ANSWER GETS BRAINLLEST If you spin the spinner below 20 times, which of the following outcomes are reasonable? Selec

t all that apply. A) Lands on blue 18 times B) Lands on red 6 times C) Lands on yellow 4 times D) Lands on blue 4 times

Mathematics
2 answers:
erik [133]3 years ago
7 0

Answer:

B) Lands on red 6 times

C) Lands on yellow 4 times

Step-by-step explanation:

The rest are way too much/too little for their predicted probability.

Tatiana [17]3 years ago
7 0

Answer:

B C

Step-by-step explanation:

If u spin the spinner below 20 times it will land on red around 4 times.

6 is only 2 greater than 4 so it works.

Same thing with yellow.

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That’s a hard one there
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Evaluate the expression when g=4 and h= 7.<br> 56<br> -9<br> h
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Answer:

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Step-by-step explanation:

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• Mother has 40 guavas to share among her daughters, Naomi got 3
scoray [572]

Answer:

Step-by-step explanation:

As given Total Guavas  = 40 ,

3 daughters,

Let   share of Sharon = x

then by given condition ,

Naomi share = x+3                  (3 more then Sharon

Also Kassie share = x+3+10 =x+13     (10 more then Naomi

Now

total shared = total

Naomi + Sharon + Kassie = 40

(x+3) +(x) +(x+13) =40

adding like terms with like we get ,

3x+16=40

3x=40-16

3x=24

x= 24/3

x= 8  

hence Sharon share = x =8

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3 years ago
Read 2 more answers
Sin x = cos 53∘ What is the value of x?
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Try this solution:
according to the rule sin(90-y)=cos(y), it is possible to re-write the given equation:
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Then substituting the value y=53° into the equation 90-y=x, it is possible to find the 'x': 90-53=37°.

answer: 37°
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3 years ago
have two one-quart jars; the first is filled with water, and the second is empty. I pour half of the water in the first jar into
vfiekz [6]

Answer:

water in quarts is in the first jar after 10th pour = 12/11

Step-by-step explanation:

Let X represent first jar and Y represents second jar.

  • have two one-quart jars; the first is filled with water, and the second is empty

Lets give the initial value of 2 to the first jar which is filled with water. Lets say there are two liters of water in first jar.

Lets give the initial value of 0 to the second as it is empty.

So before any pour, the values are:

X: 2

Y: 0

  • pour half of the water in the first jar into the second

After first pour the value of jar X becomes:

Previously it was 2.

Now half of water is taken i.e. half of 2

2 - 1 = 1

So X = 1

The value of jar Y becomes:

The half from jar X is added to second jar Y which was 0:

After first pour the value of jar Y becomes:

0 + 1 = 1

Y = 1

  • a third of the water in the second jar into the first

After second pour the value of jar X becomes:

Previously it was 1.

Now third of the water in second jar Y is added to jar X

1 + 1/3

=  (3 + 1)/3

= 4/3

X = 4/3

After second pour the value of jar Y becomes:

Previously it was 1.

Now third of the water in Y jar is taken and added to jar X so,

1 - 1/3

=  (3 - 1)/3

= 2/3

Y = 2/3

  • a fourth of the water in the first jar into the second

After third pour the value of jar X becomes:

Previously it was 4/3.

Now fourth of the water in the first jar X is taken and is added to jar Y

= 3/4 * (4/3)

= 1

X = 1

After third pour the value of jar Y becomes:

Previously it was 2/3

Now fourth of the water in the second jar X is added to jar Y

= 2/3 + 1/4*(4/3)

= 2/3 + 4/12

= 1

Y = 1

  • a fifth of the water in the second jar into the first

After fourth pour the value of jar X becomes:

Previously it was 1

Now fifth of the water in second jar Y is added to jar X

= 1 + 1/5*(1)

= 1 + 1/5

=  (5+1) / 5

= 6/5

X = 6/5

After fourth pour the value of jar Y becomes:

Previously it was 1.

Now fifth of the water in Y jar is taken and added to jar X so,

= 1 - 1/5

= (5 - 1)  / 5

= 4/5

Y = 4/5

  • a sixth of the water in the first jar into the second

After fifth pour the value of jar X becomes:

Previously it was 6/5

Now sixth of the water in the first jar X is taken and is added to jar Y

5/6 * (6/5)

= 1

X = 1

After fifth pour the value of jar Y becomes:

Previously it was 4/5

Now sixth of the water in the first jar X is taken and is added to jar Y

= 4/5 + 1/6 (6/5)

= 4/5 + 1/5

= (4+1) /5

= 5/5

= 1

Y = 1

  • a seventh of the water in the second jar into the first

After sixth pour the value of jar X becomes:

Previously it was 1

Now seventh of the water in second jar Y is added to jar X

= 1 + 1/7*(1)

= 1 + 1/7

=  (7+1) / 7

= 8/7

X = 8/7

After sixth pour the value of jar Y becomes:

Previously it was 1.

Now seventh of the water in Y jar is taken and added to jar X so,

= 1 - 1/7

=  (7-1) / 7

= 6/7

Y = 6/7

  • a eighth of the water in the first jar into the second

After seventh pour the value of jar X becomes:

Previously it was 8/7

Now eighth of the water in the first jar X is taken and is added to jar Y

7/8* (8/7)

= 1

X = 1

After seventh pour the value of jar Y becomes:

Previously it was 6/7

Now eighth of the water in the first jar X is taken and is added to jar Y

= 6/7 + 1/8 (8/7)

= 6/7 + 1/7

= 7/7

= 1

Y = 1

  • a ninth of the water in the second jar into the first

After eighth pour the value of jar X becomes:

Previously it was 1

Now ninth of the water in second jar Y is added to jar X

= 1 + 1/9*(1)

= 1 + 1/9

=  (9+1) / 9

= 10/9

X = 10/9

After eighth pour the value of jar Y becomes:

Previously it was 1.

Now ninth of the water in Y jar is taken and added to jar X so,

= 1 - 1/9

=  (9-1) / 9

= 8/9

Y = 8/9

  • a tenth of the water in the first jar into the second

After ninth pour the value of jar X becomes:

Previously it was 10/9

Now tenth of the water in the first jar X is taken and is added to jar Y

9/10* (10/9)

= 1

X = 1

After ninth pour the value of jar Y becomes:

Previously it was 8/9

Now tenth of the water in the first jar X is taken and is added to jar Y

= 8/9 + 1/10 (10/9)

= 8/9 + 1/9

= 9/9

= 1

Y = 1

  • a eleventh of the water in the second jar into the first

After tenth pour the value of jar X becomes:

Previously it was 1

Now eleventh of the water in second jar Y is added to jar X

= 1 + 1/11*(1)

= 1 + 1/11

= (11 + 1) / 11

= 12/11

X = 12/11

After tenth pour the value of jar Y becomes:

Previously it was 1.

Now eleventh of the water in Y jar is taken and added to jar X so,

= 1 - 1/11

=  (11-1) / 11

= 10/11

Y = 10/11

3 0
3 years ago
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