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3 years ago

Write a real word problem that compares 0.4 and 0.29. Then solve the problem.

Mathematics

1 answer:

sesenic [268]3 years ago

4 0

If a piece of candy costs .29$ and I have .40$ I will have .11$. Hope it helped

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Can you please help me i need help!!

Goshia [24]

Answer:

A. The fraction 1 $\frac{1}{6}$ would negative, since she is cutting off/losing hair rather than gaining it.

B. (this is just an improper fraction converted to have the same denominator as the second fraction, and is equivalent to -1 $\frac{1}{6}$ )--> - $\frac{14}{12}$ + $\frac{5}{12}$ = - $\frac{9}{12}$ or - $\frac{3}{4}$

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3 years ago

B) If you were to roll the dice 100 times, how often would you expect to get doubles?

Novosadov [1.4K]

If you're rolling a single cube, then ...

-- There are 6 possible out comes.

-- One of them (5) is a multiple of 5 .

-- The probability of rolling it is (1/6) = 16-2/3 % .

If everything is acting perfectly random, then ion 100 rolls,

you'd expect to succeed 16 or 17 times.

If you're rolling a pair of cubes, then ...

-- There are (6 x 6) = 36 possible outcomes.

-- There are seven ways to get a multiple of 5.

(1+4), (4+1), (2+3), (3+2), (4+6), (6+4), (5+5)

-- The probability of rolling one of those is

(7/36) = 19-4/9 % .

If everything is acting perfectly random, then in 100 rolls,

you'd expect to succeed 19 or 20 times.

6 0

3 years ago

What is the distance between -7 and 21 on a number line?

pshichka [43]

Answer:

The distance is 28

5 0

3 years ago

Read 2 more answers

the weight of an object on mars varies directly as the weight of the object on earth a 90 pound object on earth weighs 34 pounds

Mariulka [41]

We have been given that the weight of an object on the mars varies directly as the weight of the object on earth.

Let y be weight of an object on Mars and x be weight of an object on Earth.

$\\
y=kx$

We have been given that the weight of an object on Mars is 34 pounds and weight of the same object on Earth is 90 pounds.

Putting the given values in equation we get

$\\
34=k(90)\\
\\
k=\frac{34}{90}=\frac{17}{45}$

Therefore, our equation now becomes

$\\
y= \frac{17}{45}x$

Now we have been given that the weight of an object on Earth is 135 pounds and we have to find the weight of the same object on Mars,

We have been given that $\\
x=135$

Putting the given values in our equation we get

$\\
y= \frac{17}{45}(135)=\frac{17\cdot 135}{45}=17\cdot 3 = 51$

Therefore weight of the object on Mars will be 51 pounds.

7 0

4 years ago

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Find the area of trapezium

faltersainse [42]

Answer:

$\frac{1}{2} \times b \times h$

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3 years ago

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