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3 years ago

5

A box contains two purple marbles nine orange marbles and seven yellow marbles if you pick two without looking what is the proba

bility that both will be orange

1 answer:

maw [93]3 years ago

8 0

Answer:

1/4

Step-by-step explanation:

First, add up the marbles:

2 + 9 + 7 = 18

Multiply by 2 because you pick two:

18 × 2 = 36

Simplify:

9/36 = 3/12 = 1/4

Hope this helped.

You might be interested in

Write the expression as a single term.<br> 3ln(x+2) - ln(x - 1)

Dafna1 [17]

Answer:

ln [(x + 2)^3 ]/(x-1)

Step-by-step explanation:

3ln(x+2) - ln(x - 1) = ln (x + 2)^3 - ln (x-1)

ln (x + 2)^3 - ln (x-1) = ln [(x + 2)^3 ]/(x-1)

7 0

3 years ago

Miguel has one dollar in change. He has twice as many nickels as dimes. He has 14 more pennies than quarters. What coins does he

marin [14]

D=dimes
n=niclkes
p=pennies
q=quarters

1 dollar (100 cents)
2 times as many nickles as dimes
n=2d

14 more pennies than quarters
p+14=q

and total is 1 dollar so
25q+10d+5n+p=100
ok so we must use logic
the max number of quarters in a dolar is 4
but there are pennies so max is 3
so q=1,2, or 3
therefor p=15,16 or, 17

also, the rest of te coins are only multipules of 5 so the sum amount of pennies and quartes must add to a multiplule of 5

if q=1 and p=15, total is 40 cents
if q=2 and p=16, total is 66 cents
if q=3 and p=17, total is 92 cents
only 40 cents is a multiplule of 10

1 quarter and 15 pennies
now we are having
40 cents is pennies and quartesrs
therfor 100-40=60 cents is nickles and dimes
60=5n+10d
n=2d
sub
60=5(2d)+10d
60=10d+10d
60=20d
divide both sides by 20
3=d
sub
n=2d
n=2(3)
n=6

6 nickles
3 dimes

answer is
1 quarter
3 dimes
6 nickles
15 pennies

4 0

3 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago

xz_007 [3.2K]

Answer:

If x = 3y and 3y = x, then y = 1/3x.

3 0

3 years ago

How does the order of operations help solve one and two step equations

egoroff_w [7]

If you go from right to left, or left to right, you won't get the right answers. By using order of operations, you will get a more accurate and correct answer.

Hope this helps :)

4 0

3 years ago

Read 2 more answers