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3 years ago

What is q? For this 30 and 60, 90 angle?

Mathematics

1 answer:

Effectus [21]3 years ago

8 0

30 60 90 triangle
ratio of short leg : long leg = a : a√3

Given long leg = 9 cm
so
short leg = 9 / √3
= 3√3
= √27
= 27^(1/2) <..........(radical form)

Hope it helps.

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1) 14, 16, 18, 20, 22, ...<br> A) 29, 32, 35 B) 24, 26, 28<br> C) 34, 38, 42 D) 28, 31, 34

sergij07 [2.7K]

Answer:

What should we do the question pls

4 0

2 years ago

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

7 0

3 years ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

2 years ago

Question 4

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

Given

See attachment for chart

Required

Number of off days

To do this, we simply calculate the expected value of the chart.

This is calculated as:

$E(x) = \sum x * f$

Where

x = days

f = chances

So, we have:

$E(x) = 0* 0.3 + 1 *0.2 + 2 * 0.25 + 3 * 0.2 + 4 * 0.02 + 5 * 0.01 + 6 * 0.01 + 7 * 0.01$

$E(x) = 1.56$

$E(x) \approx 2$

3 0

2 years ago

Mrs. Martini is 6 years more than 3 times as old as Tony. Eight years ago, she

zalisa [80]

Mrs Martini age is 42 years and Tony age is 12 years

Step-by-step explanation:

The given is:

Mrs. Martini is 6 years more than 3 times as old as Tony
Eight years ago, she was 2 years less than 9 times as old as he was then

Assume that Tony is x years old now

∵ Mrs. Martini is 6 years more than 3 times as old as Tony

∵ Tony is x years old

- Multiply the age of Tony by 3 and add 6 to the product

∴ The age of Mrs Martini = 3 x + 6

8 years ago

- Subtract 8 from the age of each one

Tony age = (x - 8)

Mrs Martini age = (3 x + 6) - 8 = 3 x + 6 - 8 = 3 x - 2

∵ Eight years ago, Mrs Martini was 2 years less than 9 times as old

as Tony was then

- Multiply Tony age by 9 and subtract 2 from the product

∵ Tony age from 8 years ago = x - 8

∵ Mrs Martini age from 8 years ago = 3 x - 2

∴ 9(x - 8) - 2 = 3 x - 2

- Simplify the left hand side

∴ 9 x - 72 - 2 = 3 x - 2

- Add like terms

∴ 9 x - 74 = 3 x - 2

- Subtract 3 x from both sides

∴ 6 x - 74 = -2

- Add 74 to both sides

∴ 6 x = 72

- Divide both sides by 6

∴ x = 12

∵ x represents the age of Tony

∴ Tony is 12 years old

∴ 3 x + 6 is the age of Mrs Martini

- Substitute x by 12

∵ 3(12) + 6 = 36 + 6 = 42

∴ Mrs Martini is 42 years old

Mrs Martini age is 42 years and Tony age is 12 years

Learn more:

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6 0

2 years ago