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Aneli [31]
3 years ago
13

Maggie had $4 in her wallet. If this is 16% of her monthly allowance, wha is her monthly allowance?

Mathematics
2 answers:
lara [203]3 years ago
8 0

Answer:

<u>$25</u>

Step-by-step explanation:

Let her monthly allowance be M.

=> M x 16% = 4

=> M = 4 x 25 / 4

=> M = 25

Maggie's monthly allowance is <u>$25</u>

lukranit [14]3 years ago
8 0
$4 = 16% of her monthly allowance

4 divided by 16 = $0.25

$0.25 = 1% of her monthly allowance

$0.25 • 100 = $25

Maggie’s monthly allowance is $25.
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Q6) Renee buys 5kg of sweets to sell. She pays £10 for the sweets. Renee puts all the sweets into bags. She puts 250g of sweets
My name is Ann [436]

Answer:

30% percent profit

Step-by-step explanation:

100 pence in a pound.

1000 grams in a kilogram.

She bought 5000g of sweets.

She pays 10×100=1000 pence.

She put them into 5000÷250=20 bags.

She sells the bags for 65 pence each.

She sells them all for 20×65=1300 pence.

She gets a profit of 300 pence or 3 pounds.

She gets a percent profit of 30%.

3 0
3 years ago
One is 2% of what number?
Gnesinka [82]
I believe the answer is One is 2% of 50
7 0
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Whats slope intercept form for 2x- 5y=10
damaskus [11]

Answer:

y=-2/5x-2

Step-by-step explanation:

First, you put x to the other side of the equation so you get y alone, getting the equation of -5y=-2x+10. In order to just have y on one side, you need to divide the whole equation by -5. This leads to the final answer of y=-2/5x-2.

3 0
3 years ago
Write an expression to represent:<br> Nine minus the quotient of two and a number x
olga_2 [115]

Answer:9-(2/1x)

Step-by-step explanation:

6 0
4 years ago
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An urn contains p black balls, q white balls, and r red balls; and n balls are chosen without replacement. a. Find the joint dis
Irina-Kira [14]

The joint distribution of the numbers of the three colours in the sample without replacement is:

\mathbf{P(A=a,B=b,C=c) = \dfrac{\Big( ^{p}_{  A} \Big)  \Big( ^{q}_{  B} \Big)  \Big( ^{r}_{  C} \Big) }{ \mathbf{\Big( ^{p+q+r}_{ \ \ \ n} \Big) }   }\  \ \ \ where; n = A+B+C}

Let consider A, B, and C to denote the three variables that are black, white, and red balls in the sample.

i.e.

  • n = A + B + C

Now, the numbers of ways 'n' balls are chosen without replacement in an urn that comprises of p black balls, q white balls, and r red balls can be computed as follows:

\mathbf{\implies \Big( ^{p+q+r}_{ \ \ \ n} \Big) }

Now, the number of ways whereby A black balls can be chosen from p black balls is expressed as:

\mathbf{\implies \Big( ^{p}_{  A} \Big) }

The number of ways whereby B white balls can be chosen from q white balls is expressed as:

\mathbf{\implies \Big( ^{q}_{  B} \Big) }

The number of ways whereby C red balls can be chosen from r red balls is expressed as:

\mathbf{\implies \Big( ^{r}_{  C} \Big) }

Therefore, we can conclude that the joint distribution of the numbers of the three colours in the sample without replacement is:

\mathbf{P(A=a,B=b,C=c) = \dfrac{\Big( ^{p}_{  A} \Big)  \Big( ^{q}_{  B} \Big)  \Big( ^{r}_{  C} \Big) }{ \mathbf{\Big( ^{p+q+r}_{ \ \ \ n} \Big) }   }\  \ \ \ where; n = A+B+C}

Learn more about joint distribution here:

brainly.com/question/17283589?referrer=searchResults

4 0
3 years ago
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