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2 years ago

9

There is a bag filled with 4 blue, 3 red and 5 green marbles. A marble is taken at random from the bag, the colour is noted and

then it is replaced. Another marble is taken at random. What is the probability of exactly 1 red?

1 answer:

Ivanshal [37]2 years ago

8 0

Answer:

There is an 25% chance of getting a red marble

Step-by-step explanation:

4+3+5=12

3/12=.25

put the decimal over two spaces and you get 25%

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Eliot and his friends ate 3/8 of a pie. If the pie was cut into eight pieces ,how much pie is left over

topjm [15]

Answer:

5 pieces of pie are now left over

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2 years ago

The graph below shows a proportional relationship between x and y

Snezhnost [94]

Answer:

You get the answer by seeing if the line on the graph passes through the origin and if it does you see how many there is for every 1 thing.

Step-by-step explanation:

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3 years ago

Layla went shopping for camping equipment. After looking at several different stores, she bought a tent for $163.63 and a sleepi

Ghella [55]

Answer:

$297.09

Step-by-step explanation:

To answer this question, just add:

163.63 + 133.46 = 297.09

So, Layla spent $297.09 in all

Hope this helps :)

5 0

2 years ago

Can somebody please clearly (and correctly) explain how to successfully work out this question?

STatiana [176]

Answer:

No

Step-by-step explanation:

To calculate the number of tiles needed

divide 4m by 0.2m for row of tiles ⇒ 20 tiles per row

divide 3m by 0.2m for column of tiles ⇒ 15 tiles per column

number of tiles = 20 × 15 = 300

number of packs = 300 ÷ 10 = 30

cost = 30 × £34.99 = £1049.70

Since she has £1000 to spend and £1049.70 > £1000

She does not have enough to cover the wall

6 0

2 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

2 years ago