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3 years ago

Will y’all help me on 10

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

4 0

Answer:

The answer is 11

likoan [24]3 years ago

4 0

Answer:11

Step-by-step explanation:

She needs 100 candles but has already 15,so she will need another 100-15 candles to get to 100.

100-15 = 85

Therefore, she needs to buy 85 candles.

However,each box has 8 candles,so she will need al least 11 boxes to get a number bigger than 85.

8*11 =88

She will also have 3 spare candles as 88- 85 =3

You might be interested in

Refer to the accompanying data set of mean drive-through service times at dinner in seconds at two fast food restaurants. const

Anton [14]

Step-by-step explanation:

Assuming the data is as shown, restaurant X has a mean service time of 180.56, with a standard deviation of 62.6.

The standard error is SE = s/√n = 62.6/√50 = 8.85.

At 95% confidence, the critical value is z = 1.960.

Therefore, the confidence interval is:

180.56 ± 1.960 × 8.85

180.56 ± 17.35

(163, 198)

Restaurant Y has a mean service time of 152.96, with a standard deviation of 49.2.

The standard error is SE = s/√n = 49.2/√50 = 6.96.

At 95% confidence, the critical value is z = 1.960.

Therefore, the confidence interval is:

152.96 ± 1.960 × 6.96

152.96 ± 13.64

(139, 167)

4 0

3 years ago

I WILL GIVE BRAINLIST

docker41 [41]

I’m pretty sure it’s true

8 0

3 years ago

An item is priced at $13.57. If the sales tax is 7%, what does the item cost including sales tax?

cluponka [151]

Answer:

14.52

Step-by-step explanation:

13.57 + 13.57 x 7% = 13.57 x (1+7/100) =13.57 x 107/100 = 14.52

5 0

3 years ago

I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.

masya89 [10]

Answer:

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

4 0

2 years ago

The price of an outdoor grill was marked up 15% over the previous year's model. If the previous year's model sold for $500, what

harkovskaia [24]

I thank the answer is 75 #I thank

8 0

2 years ago