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maksim [4K]
2 years ago
5

A spinner has 5 equally sized sections, 2 of which are gray and 3 of which are blue. The spinner is spun twice. What is the prob

ability that the first spin lands on gray and the second spin lands on blue ?
Mathematics
1 answer:
maria [59]2 years ago
7 0

Answer:

6/25

Step-by-step explanation:

P(gray) = 2/5

P(blue) = 3/5

P(1st gray & 2nd blue) = (2/5)*(3/5) = 6/25

As the question already indicated the sequence, so no combination (nCr) should be considered.

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A baby elephant weighs 1,400 pounds, which is 7,437 pounds less than her mother. Write an equation to express the weight of the
erica [24]
Let's say that y is the weight of the mother and x is the weight of the baby.
We know that the baby is 7437 pounds less than the mother, which means you need to add 7437 pounds to the baby's weight to get the mother's.
We can show it like this:
x+7437=y.
We know what the baby's weight is (1400 lbs), so just sub it in:
1400+7437=y
y=8837 lbs
The mom weighs 8837 pounds.
4 0
4 years ago
Determine the first four terms of the sequence in which the nth term is
Licemer1 [7]

<u>Answer:</u>

The correct answer option is: \frac{1}{3} ,\frac{1}{4} ,\frac{1}{5} ,\frac{1}{6}.

<u>Step-by-step explanation:</u>

We know that the nth term a_n for an arithmetic sequence is given by:

a_n=\frac{(n+1)!}{(n+2)!}

where n is the number of the position of the term.

We are supposed to find the first four terms of the sequence so we will substitute the values of n from 1 to 4 in the given formula to get:

1st term:

a_1=\frac{(1+1)!}{(1+2)!}=\frac{1}{3}

2nd term:

a_2=\frac{(2+1)!}{(2+2)!}=\frac{1}{4}

3rd term:

a_3=\frac{(3+1)!}{(3+2)!}=\frac{1}{5}

4th term:

a_4=\frac{(4+1)!}{(4+2)!}=\frac{1}{6}

8 0
3 years ago
NEED ANSWER ASAP REALLY IMPORTANT i will take any of the answers
dedylja [7]

1. \: MC  \: is \:  tangent \:  to \:  (B)\Rightarrow \widehat{BCM}=90° \\ 2. \: MC  \: and \: MZ \: are \:  tangent  \: to  \: (B)\Rightarrow MC = MZ\Leftrightarrow 5x - 9 = x + 7\Leftrightarrow x = 4 \\ 3. \: BM =  \sqrt{ {BC}^{2} + {MC}^{2}  }  =  \sqrt{ {5}^{2} +  {12}^{2}  }  = 13 \\ \Rightarrow EM = BM - BE = BM - BC = 13 - 5 = 8 \\ 4. \:  \tan(60°) = \frac{MC}{BC}  =  \frac{13 \sqrt{3}  }{ BC}\Leftrightarrow BC =  \frac{13 \sqrt{3} }{\tan(60°)} =   \frac{13 \sqrt{3} }{ \sqrt{3} }  = 13 \\ 5. \: MC =  \sqrt{ {BM}^{2}  -  {BC}^{2} }  =  \sqrt{ {20}^{2}  -  {12}^{2} }  = 16 \\ 6. \:BZ =  \sqrt{ {BM}^{2}  -  {MZ}^{2} } =  \sqrt{ {25}^{2}  -  {20}^{2} }  = 15 \\ \Rightarrow XE=BX+BE=BZ+BZ=15+15=30

7 0
3 years ago
Solve for w.−(14w+8)+8=3
Stella [2.4K]
= -(14w + 8) + 8 = 3
= -14w - 8 + 8 = 3
= -14w = 3
w = -3/14

In short, Your Answer would be -3/14 

Hope this helps!
6 0
3 years ago
Read 2 more answers
Which of the following patterns or images could best be described by a translation?
babunello [35]
Translations are all about shifting the same thing.
A is more of a rotation
C is more of an enlargement
D is a mere reflection of what's on the other side.

The pattern or image that could best be described by a translation is (B) <span>seats on a school bus</span>
6 0
3 years ago
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