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2 years ago

How do you work out Y/5 = 10

Mathematics

1 answer:

Misha Larkins [42]2 years ago

4 0

Answer:

Step-by-step explanation:

y/5=10

x everything by 5 to get y on its own

y=50

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Read 2 more answers

Find the probability for the experiment of tossing a coin three times. Use the sample space S = {HHH, HHT, HTH, HTT, THH, THT, T

myrzilka [38]

Answer:

1) 0.375

2) 0.375

3) 0.5

4) 0.5

5) 0.875

6) 0.5

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

$\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

$\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375$

2. The probability of getting exactly two tails

P(Exactly two tail)

Favorable outcomes ={ HTT,THT, TTH}

$\text{P(Exactly two tail)} = \dfrac{3}{8} = 0.375$

3. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

$\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5$

4. The probability of getting a tail on the last toss

P(tail on the last toss)

Favorable outcomes ={HHT,HTT,THT,TTT}

$\text{P(tail on the last toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5$

5. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

$\text{P(at least one head)} = \dfrac{7}{8} = 0.875$

6. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

$\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5$

3 0

3 years ago

Melissa has started training for a race. The first time she trains, she bikes 5 miles. Each subsequent time she trains, she bike

joja [24]

9514 1404 393

Answer:

A. d = ∑[k=1,n] (4.5+0.5k)

B. 9

Step-by-step explanation:

A) The series has first term 5 and common difference 0.5. The general term is given by ...

an = a1 +d(n -1)

an = 5 + 0.5(n -1) . . . . substitute numbers for this series

an = 4.5 +0.5n . . . . . . simplify

The sum is the sum of these terms:

$\displaystyle \boxed{d_n=\sum_{k=1}^n(4.5+0.5k)}$

B) The sum of terms of an arithmetic series is the average of the first and last, multiplied by the number of terms. We want to solve for n to make the sum greater than 60.

$\displaystyle d_n=\frac{n}{2}(5 +4.5 +0.5n)\\\\\dfrac{n}{2}(9.5+0.5n)>60\\\\n(19+n)>240\qquad\text{multiply by 4}\\\\n^2+19n+9.5^2>240+9.5^2\qquad\text{complete the square}\\\\n+9.5>\sqrt{330.25}\qquad\text{square root}\\\\n>18.17-9.5=8.67\qquad\text{positive root, subtract 9.5}$