What does 3,146 round to nearest thousand

Polly buys 14 cupcakes for a party. The bakery puts them into boxes that hold 4 cupcakes each.

Temka [501]

Answer:

Part a) The number of boxes needed will be 4

part b) The fraction of the box that is empty is $\frac{1}{2}$

Part c) Are needed 2 cupcakes to fill the last box

Step-by-step explanation:

Part a) How many boxes will be needed for Polly to bring all the cupcakes to the party?

To find out the number of boxes needed, divide the total number of cupcakes by 4 (maximum number of cupcakes that fit in each box).

so

$\frac{14}{4}=3.5\ box$

Round up

The number of boxes needed will be 4

Part b) If the bakery completely fills as many boxes as possible, what fraction of the last box is empty?

we know that

The maximum number of cupcakes per box is 4

The first three boxes have 4 cupcakes

The last box has

$14-3(4)=2\ cupcakes$

The fraction of the box that is empty is equal to the number of cupcakes missing to fill the box divided by the cupcake capacity of the box

$\frac{2}{4}=\frac{1}{2}$

Part c) How many more cupcakes are needed to fill this box

Subtract the number of cupcakes in the box from the cupcake capacity of the box

The last box has 2 cupcakes

The cupcake capacity of the box is 4

so

$4-2=2\ cupcakes$

therefore

Are needed 2 cupcakes to fill the last box

8 0

3 years ago

Is the relationship between the number of movie tickets and the total cost of the tickets proportional?

almond37 [142]

Yes, the proportion is -6.

5 0

3 years ago

PLZ NO FILES I really need help

Lelechka [254]

53 is the answer

Explanation

Yes

8 0

3 years ago

Read 2 more answers

If a 10 symbol sequence is sent through the channel,what is the probability that up to 3 symbols are in error out of the 10trans

lina2011 [118]

Answer: 0.171887

Step-by-step explanation:

Given that S0 and S1 are binary symbol of equal probabilty;

P(S0) = P(S1) = 0.5

This probability is a Binomial random variable of sequence Sn, where Sn counting the number of success in a repeated trials.

P(Sn =X) = nCx p^x (1-p)^(n-1)

Pr(at most 3) = P(0<= x <=3) = P(X=0) + 0) + P(X=1) + P(X=2) + P(X=3)

Since there are only 2 values that occur in sequence 0 and 1 ( or S0 and S1).Let the distribution be given by the sequence (0111111111),(1011111111),(11011111111),...(1111111110) for Sn= 1 is the sequence for 1 error.

10C0, 10C1, 10C2, and 10C3 is the number of sequences in value for X= 0, 1, 2, 3 having value 0 and others are 1. Let the success be p(S0)=0.5 and p(S1)= 0.5

P(0<= X <=3) = 10C0 × (0.5)^0 × (0.5)^10 + 10C1× (0.5)¹ × (0.5)^9 + 10C2 × (0.5)² × (0.5)^8 + 10C3(0.5)³(0.5)^7

= 1 × (0.5)^10 + 10 × (0.5)^10 + 45 × (0.5)^10 + 120 × (0.5)^10

=0.000977 + 0.00977 + 0.04395 + 0.11719

= 0.171887

3 0

3 years ago

How many thousands of blue crayons are in 327,412

bearhunter [10]

Answer:327412000

Step-by-step explanation:

8 0

4 years ago