Ask question

Ask question

All categories

Law Incorporation [45]

3 years ago

10

It's spaghetti night at Bonnie's house! She has 8 ounces of freshly grated cheese to sprinkle evenly over 5 bowls of spaghetti.

How much cheese will be in each bowl?

2 answers:

Alecsey [184]3 years ago

8 0

Answer:

1.6

Step-by-step explanation:

Divide 8 by 5.

Hunter-Best [27]3 years ago

3 0

Step-by-step explanation:

Answer= 1.6 ounces

8 ounces/ 5 bowls= 1.6 ounces in each bowl

You might be interested in

How do i make my grades go up quickly- yes in all subjects plz help me

sveticcg [70]

Answer:

Step-by-step explanation:

A way to make your grades go higher is by asking your teacher for extra credits. If you fail some quizzes ask your teacher to re-take it to get a better grade, another way is ask your teacher for assistance on assignments.

6 0

3 years ago

What is the exact volume of the sphere? the radius is 4.8

Allisa [31]

V = 4/3πr³
V= 4/3π(4.8)³
V= 4/3π(110.592)
V=147.456π

5 0

3 years ago

The lifespan of a 60-watt lightbulb produced by a company is normally distributed with a mean of 1450 hours and a standard devia

Flura [38]

To evaluate the probability that the lifespan will be between 1440 and 1465 hours will be given by:
P(1440<x<1465)
using the z-score formula we obtain:
z=(x-μ)/σ
where:
μ=1450
σ=8.5
hence
when x=1440
z=(1440-1450)/8.5
z=-1.18
P(z<-1.18)=0.1190

when x=1465
z=(1465-1450)/8.5
z=1.77
P(z<1.77)=0.9625

hence:
P(1440<x<1465)
=0.9625-0.1180
=0.8445

8 0

3 years ago

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

cestrela7 [59]

Answer:

(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

Given

Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

$Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}$

$n(Head) = 32$

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

$n(Dice) = 30$

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

<em>From the above calculations: </em> $0.267 > 0.03125$ <em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

$Median = 16.5th$

This means that the median is the mean of the 16th and the 17th item

So,

$Median = \frac{3+2}{2}$

$Median = \frac{5}{2}$

$Median = 2.5$

For person B

$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

$Median = \frac{31}{2}th$

$Median = 15.5th$

This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

So, we have:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

$P(5\ or\ 6) = \frac{18}{30}$

$P(5\ or\ 6) = 0.60$

$P(5\ or\ 6) = 60\%$

Question 13: Person with higher probability of 3 or more

Person A

$n(3\ or\ more) = 16$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}$

$P(3\ or\ more) = \frac{16}{32}$

$P(3\ or\ more) = 0.50$

$P(3\ or\ more) = 50\%$

Person B

$n(3\ or\ more) = 28$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}$

$P(3\ or\ more) = \frac{28}{30}$

$P(3\ or\ more) = 0.933$

$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

7 0

3 years ago

Does -2 (6 - 3x) = -12 + 6x have any solutions?

Sedbober [7]

First to answer

Answer:

This equation has Infinite solutions! or should i say that All real numbers are solutions.

Let me know if it is correct!

7 0

3 years ago

Read 2 more answers