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3 years ago

Help me with this maths literacy

Mathematics

1 answer:

astraxan [27]3 years ago

6 0

1.6.1. The strongest is British Pound ( 1 BP= R 23.156548) and the weakest is Japanese Yen ( 1 Yen=R 0.144882)
1.6.2. It means that if South African citizens wants to exchange their money they can get less British Pounds than other currency and more Japanese Yen.
1.6.3 Total:
R 42000 + R 56 000 + R 42000 = R 140000
In Australian Dollars:
140000 : 11.518501= AUD 12,154.36
1.6.4.
One hundred and forty thousand South African Rand
Twelve thousand one hundred fifty four point thirty six Australian Dollars.

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What’s greater 14/7 or 2/4

aleksandr82 [10.1K]

I think 2/4 but I'm not 100% sure sry

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3 years ago

How many 4.2 pound bags of mulch do we need to cover 60 square yards if each bag can cover 1.25 square yards?

Talja [164]

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3 years ago

Find five numbers so that the mean, median, mode and range is 4.

Nuetrik [128]

Answer:

Let’s break this apart

Well we know the median has to be 4

Since it’s 5 numbers the middle number has to be 4 since its the median.

Let’s put in what we know.

a, b, 4, d, e

Constraints:

There has to be more then 1 “4”.

e-a = 4

So using that information lets solve since the possibility is almost endless

SO lets make e 5 and a 1.

1, b, 4, d, 5

There has to be more then 1 4 so lets put that as b.

We can solve for the last remaining digits.

1+ 4 + 4 + d + 5 / 5 = 4

14 + d /5 = 4

2.2 + d = 4

1.8 = d

So now if we put in order and replace b with 1.8 and make d as the previous “b” as 4.

1, 1.8, 4, 4, 5

Thats your 5 numbers right there.

Check:

Mode is 4: yes!

Range is 4: 5-1 = 4 yes!

Median is 4: yes!

Mean is 4: 1 + 1.8 + 4 + 4 +5 / 5 = 4 yes!

Every thing checks out.

There could be a lot of possibilities.

For example take this wrong one

Lets make the same exact thing except change the a and the e.

THis is what we have,

a, b, 4, d, e

Lets make e and a as 6 and 2. 6-2 is still 4 so its possible.

2, b, 4, d, 6

And of course we need more then 1 4 so lets make d 4.

2, b, 4, 4, 6

Now solve for b in the mean.

2 + b + 4 + 4 + 6 / 5 = 4

16 +b /5 = 4

3.2 + b = 4

Solve

B = 0.8

This doesn’t work cause the median and the range has constraint here…

When doing a median, it has to be in ORDER.

2, 0.8, 4, 4 , 6 isn’t in order

ANd even when put in order.

0.8, 2, 4, 4, 6

THe range has the constraint here becuase 6 - 0.8 isn’t 4.

8 0

3 years ago

What is 492.6 ÷ 48 UIFEWFHUWJKFUJKCJUKWRFGWUIKEFHKJWJF

lutik1710 [3]

Answer:

10.2625

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

6 0

3 years ago