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guajiro [1.7K]
2 years ago
5

The weight of a small bag of sugar is 2.265 kg. the student is baking cakes for a college scholarship fundraiser. it takes 100 g

rams of sugar to make one small cake. how many cakes can be made from a single bag of sugar?
Mathematics
1 answer:
wel2 years ago
6 0
2.265 KG = 2265 grams/100 = 22.65 cakes
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In 2013, the national debt was
Reika [66]

Answer:

You can use a calculator for the decimal operations, but practice some by hand because on the quiz and the test you will not have a calculator.

Step-by-step explanation:

7 0
2 years ago
Calculate a23 for the product of the following matrices
s344n2d4d5 [400]

If A is the matrix product

A=\underbrace{\begin{pmatrix}1&2\\4&3\end{pmatrix}}_{M_1}\underbrace{\begin{pmatrix}6&1&5\\7&3&4\end{pmatrix}}_{M_2}

and a_{2,3} is the entry of A in the second row and third column, then

a_{2,3}=\begin{pmatrix}4&3\end{pmatrix}\begin{pmatrix}5\\4\end{pmatrix}=4\times5+3\times4=\boxed{32}

That is, the 2nd row, 3rd column entry of A is the product of the 2nd row of M_1 and the 3rd column of M_2.

6 0
2 years ago
Raymond bought a car for $40,000. He took a 20,000 loan from a bank at an interest rate of 15% per year for a 5 - year period. W
jenyasd209 [6]

Answer:

$35,000

Step-by-step explanation:

The amount that would be repaid = amount borrowed + interest earned on loan

interest earned on deposit can be determined by determining the simple interest

Simple interest = principal x time x interest rate

principal = the amount deposited = 20,000

Time = the duration of the deposit =5  

interest rate = the percentage on deposit that would be earned = 15

20,000 x 5 x 0.15 = $15,000

total = 20,000 + 15,000 = $35,000

3 0
1 year ago
There are 200 students in eleventh grade high school class. There are 40 students in the soccer team and 50 students in the bask
bekas [8.4K]

Answer:

P(A) = 0.2

P(B) = 0.25

P(A&B) = 0.05

P(A|B) = 0.2

P(A|B) = P(A) = 0.2

Step-by-step explanation:

P(A) is the probability that the selected student plays soccer.

Then:

P(A)=\dfrac{40}{200}=0.2

P(B) is the probability that the selected student plays basketball.

Then:

P(B)=\dfrac{50}{200}=0.25

P(A and B) is the probability that the selected student plays soccer and basketball:

P(A\&B)=\dfrac{10}{200}=0.05

P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:

P(A|B)=\dfrac{P(A\&B)}{P(B)}=\dfrac{10}{50}=0.2

P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.

4 0
2 years ago
If the test scores of a class of 35 students have a mean of 74.3 and the test scores of another class of 28 students have a mean
STALIN [3.7K]
Let the total sum of the scores of the first class of 35 students be a. The mean is 74.3 .

So 
\frac{a}{35}=74.3\\\\a=35\cdot74.3= 2600.5

Also, let the total sum of the scores of the second class of 28 students be b. The mean is 67.6 .

so 
\frac{b}{28}=67.6\\\\ b=28\cdot67.6= 1892.8



The combined group has 35+28=63 students. The sum of their scores is 

a+b=
2600.5+1892.8=4493.3


Thus, the mean of the combined group is \frac{4493.3}{63}= 71.32

6 0
2 years ago
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