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3 years ago

11

Together, Amy and Thea make up 1/4 of the midfielders on my the soccer team . How many midfielders are on the team? Show your wo

rk

1 answer:

gladu [14]3 years ago

8 0

8
2=1/4 of total
divide 2 by 1/4 and you get 8

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The population of a town increased by 15% in 2016, and decreased by 5% in 2017. If the population of the town was 60,000 in the

nikdorinn [45]

Answer:

The population of the town at the end of 2017 was 65,550.

Step-by-step explanation:

The population of the town was 60,000 in the beginning of 2016.

In 2016, the total population is increased by 15%.

$60,000\times\frac{15}{100}=9000$

Therefore the population is increased by 9000 and the population of the town at the end of 2016 was

$60,000+9000=69,000$

In 2017, the total population is decreased by 5%.

$69,000\times\frac{5}{100}=3450$

Therefore the population is decreased by 3450 and the population of the town at the end of 2017 was

$69,000-3450=65,500$

Therefore the population of the town at the end of 2017 was 65,550.

3 0

4 years ago

£3.50= %of£10 answer this question i raelly dont get it in any way

ziro4ka [17]

$\dfrac{\pounds 3.50}{\pounds10}\cdot100\%=\\
0.35\cdot100\%=\\
35\%$

3 0

3 years ago

6. Which of the following will NOT produce more

ziro4ka [17]

Answer:

i think its obtuse scalene ??

Step-by-step explanation:

i might not be right but im pretty sure

7 0

3 years ago

Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are sel

serious [3.7K]

Complete Questions:

Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are selected does not matter, from the positive integers not exceeding the given integers.

a. 40

b. 48

c. 56

d. 64

Answer:

a. 0.35

b. 0.43

c. 0.49

d. 0.54

Step-by-step explanation:

(a)

The objective is to find the probability of selecting none of the correct six integers from the positive integers not exceeding 40.

Let s be the sample space of all integer not exceeding 40.

The total number of ways to select 6 numbers from 40 is $|S| = C(40,6)$ .

Let E be the event of selecting none of the correct six integers.

The total number of ways to select the 6 incorrect numbers from 34 numbers is:

$|E| = C(34,6)$

Thus, the probability of selecting none of the correct six integers, when the order in which they are selected does rot matter is

$P(E) = \frac{|E|}{|S|}$

$= \frac{C(34, 6)}{C(40, 6)}\\\\= \frac{1344904}{3838380}\\\\=0.35$

Therefore, the probability is 0.35

Check the attached files for additionals

8 0

3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago