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4 years ago

7

The nutritional information on Xuan’s container of crackers said that it had 162 calories. If there were 24 crackers in the cont

ainer, which best describes the number of calories in each cracker?

1 answer:

Colt1911 [192]4 years ago

4 0

Answer:

B. approximately 7 calories

Step-by-step explanation:

got it right, hope you do as well

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Hal is thinking of a number. If he multiplies his number by three and then adds 4, he gets 25. What is Hal’s number?

lesya [120]

Answer:

7

Step-by-step explanation: Ok this coming from a 11 year old. if 7x3= 21 then add 4 its 25 simple

6 0

3 years ago

WILL GIVE BRAINLIEST FOR RIGHT ANSWER!!

11111nata11111 [884]

Answers:

The equation is W = 33T + 500
After 14 minutes, there are <u> 962 </u> liters of water.

=======================================================

Explanation:

T = number of minutes
33T = amount of added water in liters, since we're adding 33 liters per min.
33T+500 = adding on the initial 500 liters

This leads to the equation W = 33T + 500

------------------------

Plug in T = 14

W = 33T + 500

W = 33*14 + 500

W = 462 + 500

W = 962 liters of water are in the pond after 14 minutes

7 0

2 years ago

Please Solve this Math Proof

Leokris [45]

There is nothing here....

4 0

4 years ago

ELEN [110]

Answer:

Step-by-step explanation:

The size does not change because the shape is only moving to new coordinates not growing or shrinking.

This means the shape also does not change.

When you translate a shape in geometry it means you are changing the position. But the shape will remain the same, meaning the angles remain congruent.

8 0

3 years ago

A bag contain 3 black balls and 2 white balls.

Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

1)

a)

$Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\$

b)

$Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

c)

Probability of at least one black( means BB or BW or WB)

$=\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}$

d)

Probability of at most one black ( means WW or WB or BW)

$=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}$

2)

a) Probability both black without replacement

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

b) Probability of one black and one white

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

c) Probability of at least one black ( BB or BW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}$

d) Probability of at most one black ( BW or WW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}$

6 0

3 years ago