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2 years ago

Cookies cost $3 each, and candies cost $2 each. Tim has

Mathematics

2 answers:

riadik2000 [5.3K]2 years ago

6 0

Answer:

he can buy 5 cookies and 5 candies

Step-by-step explanation:

3+2=5

5×5=25

REY [17]2 years ago

3 0

Answer:

8x < 12y

Step-by-step explanation:

If you divide 25 by 3, you get 8.3333... (the threes go on forever), which you can round to 8, because little Timmy isn't going to buy .3 of a cookie. If you divide 25 by 2, you get 12.5, which you round down to 12. I know you normally don't round 5 down, but we are talking about food here.

Therefore, Tim can buy less cookies than candies. He can buy 8 cookies, and 12 candies.

Either way, Timmy's gonna be fat.

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Write the following expression in standered form.<br><br> -3•2•a+5(-7b)+(12•c•4)

Alex

Answer:

−6a − 35b + 48c

Step-by-step explanation:

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2 years ago

There is a puddle 1.4 cm deep in your backyard. After one minute of rain, the puddle was 1.45 cm deep. The puddle was 1.5 cm dee

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2 years ago

Consider the given geometric sequence.

Gala2k [10]

F5=(256)f1=16 hope this helps

7 0

2 years ago

Read 2 more answers

Assume that trees are subjected to different levels of carbon dioxide atmosphere with 7% of the trees in a minimal growth condit

taurus [48]

Answer: The mean and standard deviation are 567.2 and 89.88 resp.

Step-by-step explanation:

Since we have given that

For 370 parts per million = 7% = 0.07

For 440 parts per million = 10% = 0.10

For 550 parts per million = 49% = 0.49

For 670 parts per million = 34% = 0.34

So, Mean of the carbon dioxide atmosphere for these trees would be

$E[x]=370\times 0.07+440\times 0.1+550\times 0.49+670\times 0.34=567.2$

And

$E[x^2]=370^2\times 0.07+440^2\times 0.1+550^2\times 0.49+670^2\times 0.34=329794$

So, Variance would be

$Var\ x=E[x^2]-E[x]^2=329794-567.2^2=8078.16$

So, the standard deviation would be

$\sigma=\sqrt{8078.16}=89.88$

Hence, the mean and standard deviation are 567.2 and 89.88 resp.

3 0

3 years ago

The amount of polonium-210 remaining, p(t), after t days in a sample can be modeled by the exponential function p(t) = 100e−0.00

Archy [21]

Answer:

% Po lost = 100[1 - e^(-0.005t)] %; 73.0 g

Step-by-step explanation:

p(t) = 100e^(-0.005t)

Initial amount: p(0) = 100

Amount remaining: p(t) = 100e^(-0.005t)

Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]

% of Po lost = amount lost/initial amount × 100 %

= [1 - e^(-0.005t)] × 100 % = 100[1 - e^(-0.005t)] %

p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g

The mass of polonium remaining after 63 days is 73 g.

4 0

2 years ago