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3 years ago

Ian uses 4 feet of ribbon to wrap each package. How many packages can he wrap with 5.5 yards of ribbon?

1 answer:

4 0

First you want to turn yard into feet so 5.5 times 3 which makes 16.5 then divide by 4 and you get 4.125 but you can't do that so it would be 4 packages

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Add and subtract fractions<br><br><br>2/(x-2) + 3/(2-x) + 1/(x+3)

Oksanka [162]

Once you add and simplify you’re answer will be
- 5/(x+3)(x-2)

5 0

3 years ago

Emily has 12 fruits in her bowl. She has 3 apples, 5 bananas, 1 pear and 3 oranges. A fruit is selected at random. What is the p

blsea [12.9K]

Answer: The probability that the fruit is an orange or a pear is $\dfrac{1}{3}$ .

Step-by-step explanation:

Given : Emily has 12 fruits in her bowl.

She has 3 apples, 5 bananas, 1 pear and 3 oranges.

A fruit is selected at random.

P(orange) = $\dfrac{\text{Number of oranges}}{\text{Total fruits}}$

$=\dfrac{3}{12}$

P(pear)= $\dfrac{\text{Number of pears}}{\text{Total fruits}}$

$=\dfrac{1}{12}$

Since both events of selecting orange and pear are mutually exclusive , so

The probability that the fruit is an orange or a pear = P(orange) + P(pear)

$=\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}$

Therefore , the probability that the fruit is an orange or a pear is $\dfrac{1}{3}$ .

5 0

3 years ago

Point P is located at (-3,5)

Naya [18.7K]

Answer:

P' (- 7, - 5 )

Step-by-step explanation:

A translation of 3 units to the left means subtractin 3 from the x- coordinate with no change to the y- coordinate, thus

P(- 3, 5 ) → (- 3 - 4, 5 ) → (- 7, 5 )

Under a reflection in the x- axis

a point (x, y ) → (x, - y ), thus

(- 7, 5 ) → P'(- 7, - 5 )

5 0

3 years ago

Jim's backyard is a rectangle that is 15 5/6 yards long and 10 2/5 yards wide. Jim buys sods in pieces that are 1 1/3 yards long

Ne4ueva [31]

We are given

Jim's backyard:

Length is

$L=15\frac{5}{6} =\frac{95}{6} yards$

width is

$W=10\frac{2}{5} =\frac{52}{5} yards$

Since, this is rectangle

so, we can find area of rectangle

$A=L*W$

$A=(\frac{95}{6})*(\frac{52}{5})$

$A=\frac{494}{3} yard^2$

Area of one sod:

length is

$l=1\frac{1}{3} =\frac{4}{3} yards$

width is

$w=1\frac{1}{3} =\frac{4}{3} yards$

Since, it is rectangle in shape

so,

$Area=l*w$

$A=\frac{4}{3}*\frac{4}{3}$

$A=\frac{16}{9}yard^2$

Number of pieces of sod:

we can use formula

Number of pieces of sod = (area of Jim's backyard)/(area of one sod)

$N=\frac{\frac{494}{3} }{\frac{16}{9} }$

now, we can simplify it

$N=\frac{741}{8}$ pieces need ..............Answer

7 0

3 years ago

Can someone help . 8th grade math.

SVETLANKA909090 [29]

the answer should be if i did my math right is 0.25

8 0

3 years ago