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3 years ago

6

A high school soccer goalie blocked the ball from going into the goal 4 out of 5 times. If the ball kicked toward the goal 25 ti

mes,how many blocks did the goalie make ?

2 answers:

malfutka [58]3 years ago

8 0

Hello!

You have to find $\frac{4}{5}$ of 25

$\frac{4}{5} = \frac{x}{25}$

Cross multiply

25 × 4 = 100

100 ÷ 5 = 20

The goalie made <u>20 blocks</u>

miskamm [114]3 years ago

4 0

We can use a ratio expression to find the answer to this problem.

Set up the ratio.
4/5 = x/25
Multiply both sides by 25.
100/5 = x
Divide by 5.
20 = x
This means out of 25 times the ball is kicked toward the goal, the goalie will block it 20 times.

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Ivan and Tanya share £150 in the ration of 4:1

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Answer:

Money Ivan gets extra = 120 - 30 (or 3x) = 90

Step-by-step explanation:

Ivan = 4x

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Mystery Boxes: Breakout Rooms

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Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

<u>Lower half:</u>

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

<u>Upper half</u>

<u></u> $\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$ <u></u>

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

<u></u> $\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$ <u></u>

<u></u>

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

<u>Lower half:</u>

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question, the median is:

$Median = 22$ and $N = 14$

To calculate the median, we make use of:

$Median = \frac{N + 1}{2}th$

$Median = \frac{14 + 1}{2}th$

$Median = \frac{15}{2}th$

$Median = 7.5th$

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The 7th and 8th items are blanks.

However, from the question; the mode is:

$Mode = 18$

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

$7th = 18$

The 8th item is calculated as thus:

$Median = \frac{1}{2}(7th + 8th)$

$22= \frac{1}{2}(18 + 8th)$

Multiply through by 2

$44 = 18 + 8th$

$8th = 44 - 18$

$8th = 26$

The updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

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$Mean = 26$

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}$

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$26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}$

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$14 * 26= 2nd + 12th+304$

$364= 2nd + 12th+304$

This gives:

$2nd + 12th = 364 - 304$

$2nd + 12th = 60$

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$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We know that:

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$2nd = 3$

$12th = 57$

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$2nd + 12th = 60$

$3 + 57 = 60$

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$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

6 0

2 years ago

A length of chain is to be constructed by placing 36 component links end to end. The length of a link produced by a production p

Setler79 [48]

Answer:

d. 90 cm, 1.2 cm

Step-by-step explanation:

Given that:

The sample size n = 36

For the length of the link:

The mean $\overline x$ = 2.5

The standard deviation s = 0.2 cm

In a chosen 36 links, The mean and standard deviation for the length chain is as follows:

Mean = n $\overline x$

Mean = 36×2.5

Mean = 90 cm

The Standard deviation = s×√n

The Standard deviation = 0.2×√36

The Standard deviation = 0.2×6

The Standard deviation = 1.2 cm

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KonstantinChe [14]

Answer:

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Step-by-step explanation:

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