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3 years ago

A soccer team won 62% of their games. How many did they win if they payed 50 games?

Mathematics

1 answer:

Varvara68 [4.7K]3 years ago

8 0

Answer:

Step-by-step explanation:

50 x .62= 31

(games played) x the percentage won.

You could muliply 50 by 62% if your calculator has the % symbol if not you have to turn it into a decimal. To go from percent to decimal just take your percentage and move the decimal to the left twice. 62% = .62

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50 POINTS!!!

e-lub [12.9K]

10 oz of salmon is needed for 16 servings

20 oz of salmon is needed for 32 servings

50 oz of salmon needed for 80 servings

62.5 oz of salmon needed for 100 servings

<em><u>Solution:</u></em>

Given that,

One roll can be cut into 8 servings

The ingredients for one roll include sushi rice, seaweed sheets, cucumbers, cream cheese, and 5oz of smoked salmon

1 roll is 8 servings and it is made of 5 oz of smoked salmon

Which means that, 5 oz of smoked salmon is needed fo 8 servings

Write an expression for the amount of salmon needed to make s serving of sushi

$Amount\ of\ salmon = \frac{5}{8}s$

Where, s = serving

<h3><u>How much salmon is needed to make ?</u></h3><h3><u>16 servings</u></h3>

$Amount\ of\ salmon = \frac{5}{8} \times 16 = 5 \times 2 = 10$

Thus 10 oz of salmon is needed for 16 servings

<h3><u>32 servings</u></h3>

$Amount\ of\ salmon = \frac{5}{8} \times 32 = 5 \times 4 = 20\ oz$

Thus 20 oz of salmon is needed for 32 servings

<h3><u>80 servings</u></h3>

$Amount\ of\ salmon = \frac{5}{8} \times 80 = 5 \times 10 = 50\ oz$

Thus 50 oz of salmon needed for 80 servings

<h3><u>100 servings</u></h3>

$Amount\ of\ salmon = \frac{5}{8} \times 100 = 5 \times 12.5 = 62.5\ oz$

Thus 62.5 oz of salmon needed for 100 servings

4 0

3 years ago

Radius for cylinder is 4cm height 14.6 cm what is the volume

Kazeer [188]

Answer:

V≈733.88cm³

Step-by-step explanation:

Thank you

8 0

3 years ago

Multiply and simplify (4p +2) (6p-3) show work

quester [9]

Answer is 24p^2 - 6
you get this because you are suppose to use foil
4p x 6p
4p x -3
2 x 6p
2 x -3
when u do all that you get
24p^2 - 12p + 12p -6
combine like terms you get 24p^2 - 6

6 0

3 years ago

Help me with my homrwork

12345 [234]

Answer:

divide 19400000 and 51000

answer will be 380.39

6 0

2 years ago

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

Deffense [45]

Answer:

a) $17-2.26\frac{1.9}{\sqrt{10}}=15.64$

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$

So on this case the 95% confidence interval would be given by (15.64;18.36)

b) 1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Step-by-step explanation:

Part a: What is the lower bound to this confidence interval? 2 cm (round to 2 decimal places) What is the upper bound to this confidence interval? cm (round to 2 decimal places)

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

We have the following data:

$\bar x= 17$ represent the sample mean

$s = 1.9$ represent the sample deviation

n =10 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.26$

Now we have everything in order to replace into formula (1):

$17-2.26\frac{1.9}{\sqrt{10}}=15.64$

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$

So on this case the 95% confidence interval would be given by (15.64;18.36)

Part b

1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

5 0

3 years ago