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3 years ago

Sara has $50 in her wallet in $5 and $20 bills. Create a linear equation (with

Mathematics

1 answer:

77julia77 [94]3 years ago

4 0

Answer:

0 $20 bills and 10 $5 bills

1 $20 bills and 6 $5 bills

2 $20 bills and 2 $5 bills

Step-by-step explanation:

Equations

Let's set:

x=number of $5 bills

y=number of $20 bills

The total amount Sara has is given by

5x+20y

And we know it's equal to $50, thus:

5x+20y=50

Dividing by 5

x+4y=10

We would need another condition to solve for x and y, but we can determine some combinations that solve the problem.

Solving for x:

x=10-4y

Since both x and y are integers and cannot be negative:

10-4y≥0

Swapping sides:

4y≤10

Dividing by 4:

y≤2.5

Thus, y can only have the values {0,1,2}

For y=0

x=10-4*0=10

x=10

For y=1

x=10-4*1=6

x=6

For y=2

x=10-4*2=2

x=2

Thus, the possible combinations are:

0 $20 bills and 10 $5 bills

1 $20 bills and 6 $5 bills

2 $20 bills and 2 $5 bills

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Inessa05 [86]

Answer:

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(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

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