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2 years ago

5

Irene owns a bakery.For each cake,she spends 1/4 of an hour to make frosting and 2/5 hour to decorate.How many cakes can Irene b

ake in 3 1/4 hours?

1 answer:

aleksandrvk [35]2 years ago

8 0

The answer is 5 cakes.

EXPLANATION

To solve this, you work out that 1/4 of an hour is 15 mins 2/5 of an hour is 24 mins and that 3 1/4 hours is 195 mins.

Using these, we can work out that it takes 39 mins per cake (15 for frosting + 24 for decorating). When we divide 195 by this we get 5, meaning that the answer is 5 cakes.

Hope this helps!

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When Jennifer spent 3/7 of her money, she has $72 left. How much money did Jennifer start with?

weeeeeb [17]

After spending 3/7, she is left with;
4/7
Now, 4/7 is equivalent to 72
7/7 is equivalent to?
$\frac{7*72}{4}$
= 126 dollars

3 0

2 years ago

There are 210210210 students in a twelfth grade high school class. 909090 of these students have at least one sister and 1051051

BlackZzzverrR [31]

Answer:

(1) The value of P (A) is 0.4286.

(2) The value of P (B) is 0.50.

(3) The value of P (A ∩ B) is 0.2143.

(4) The the value of P (B|A) is 0.50.

(5) The events <em>A</em> and <em>B</em> are independent.

Step-by-step explanation:

The events are defined as follows:

<em>A</em> = a student in the class has a sister

<em>B</em> = a student has a brother

The information provided is:

<em>N</em> = 210

n (A) = 90

n (B) = 105

n (A ∩ B) = 45

The probability of an event <em>E</em> is the ratio of the favorable number of outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

The conditional probability of an event <em>X</em> provided that another event <em>Y</em> has already occurred is:

$P(X|Y)=\frac{P(A\cap Y)}{P(Y)}$

If the events <em>X</em> and <em>Y</em> are independent then,

$P(X|Y)=P(X)$

(1)

Compute the probability of event <em>A</em> as follows:

$P(A)=\frac{n(A)}{N}\\\\=\frac{90}{210}\\\\=0.4286$

The value of P (A) is 0.4286.

(2)

Compute the probability of event <em>B</em> as follows:

$P(B)=\frac{n(B)}{N}\\\\=\frac{105}{210}\\\\=0.50$

The value of P (B) is 0.50.

(3)

Compute the probability of event <em>A</em> and <em>B</em> as follows:

$P(A\cap B)=\frac{n(A\cap B)}{N}\\\\=\frac{45}{210}\\\\=0.2143$

The value of P (A ∩ B) is 0.2143.

(4)

Compute the probability of <em>B</em> given <em>A</em> as follows:

$P(B|A)=\frac{P(A\cap B)}{P(A)}\\\\=\frac{0.2143}{0.4286}\\\\=0.50$

The the value of P (B|A) is 0.50.

(5)

The value of P (B|A) = 0.50 = P (B).

Thus, the events <em>A</em> and <em>B</em> are independent.

6 0

2 years ago

Assume that the Poisson distribution applies and that the mean number of hurricanes in a certain area is 6.7 per year. a. Find t

igor_vitrenko [27]

Answer:

a) 10.34% probability that, in a year, there will be 4 hurricanes.

b) 3.62 years are expected to have 4 hurricanes

c) Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

6.7 per year.

This means that $\mu = 6.7$

a. Find the probability that, in a year, there will be 4 hurricanes.

This is P(X = 4).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-6.7}*(6.7)^{4}}{(4)!} = 0.1034$

10.34% probability that, in a year, there will be 4 hurricanes.

b. In a 35-year period, how many years are expected to have 4 hurricanes?

Each year, 0.1034 probability of 10 hurricanes.

In 35 years

35*0.1034 = 3.62

3.62 years are expected to have 4 hurricanes

c. How does the result from part (b) compare to a recent period of 35 years in which 3 years had 4 hurricanes? Does the Poisson distribution work well here?

Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.

6 0

2 years ago

Lola measures three buttons for a shirt she is making. one button is 1/8 inch, one is 3/8 inch, and one is 1/4 inch. the button

den301095 [7]

Answer: The first and last button will fit through the button hole.

Step-by-step explanation:

since we have given that

Size of first button is given by

$\frac{1}{8}\ inch$

Size of second button is given by

$\frac{3}{8}\ inch$

Size of third button is given by

$\frac{1}{4}\ inch$

The size of hole of the shirt is given by

$\frac{2}{6}\ inch=\frac{1}{3}\ inch$

So, the size of button must be smaller than the size of hole in the shirt to get fit .

Since we first make the denominator same , as L.C.M. of all denominators i.e.(8,8,4,and 3) is 24

So, Size of first button is given by

$\frac{1}{8}\ inch=\frac{1\times 3}{8\times 3}=\frac{3}{24}\ inch$

Similarly, Size of second button is given by

$\frac{3}{8}\ inch=\frac{3\times 3}{8\times 3}=\frac{9}{24}\ inch$

Similarly, Size of third button is given by

$\frac{1}{4}\ inch=\frac{1\times 6}{4\times 6}=\frac{6}{24}\ inch$

And, The size of hole of the shirt is given by

$\frac{1}{3}\ inch=\frac{1\times 8}{3\times 8}=\frac{8}{24}$

As we can see that

$\frac{3}{24}$

Hence, the first and last button will fit through the button hole.

7 0

2 years ago

Two number cubes are rolled to determine how a token moves on a game board. The sides of

ahrayia [7]

Answer:

D. The mathematical expectation of Option A is 1. The mathematical expectation of Option B is 1.5. Option B offers a greater likelihood of advancing to the finish line.

Step-by-step explanation:

The result of a product is odd only when the two numbers are odds.

There are 6*6 = 36 possible outcomes when two dice are rolled. Only 9 of them are a combination of two odd numbers: {1, 1} {1, 3} {1, 5} {3, 1} {3, 3} {3, 5} {5, 1} {5, 3} {5, 5}. Then 36 - 9 = 27 outcomes are even.

P(even) = 27/36 = 0.75

Option A) Mathematical expectation: 0.75*4 + 0.25*(-8) = 1

Option B) Mathematical expectation: 0.75*5 + 0.25*(-9) = 1.5

8 0

2 years ago