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2 years ago

The selling price of 3 cucumbers is $2.00

Mathematics

1 answer:

Nina [5.8K]2 years ago

4 0

Answer:

The first blank is 9 for 9 cucumbers

The second blank is 8 dollars

Step-by-step explanation:

We can use ratios to solve

3 cucumbers x cucumbers

--------------------- = ----------------------

2 dollars 6 dollars

Using cross products

3*6 = 2x

18 = 2x

Divide by 2

18/2 = 2x/2

9 =x

We need 9 cucumbers

For the third row

3 cucumbers 12 cucumbers

--------------------- = ----------------------

2 dollars y dollars

Using cross products

3*y = 2*12

3y = 24

Divide by 3

3y/3 = 24/3

y = 8

We need 8 dollars

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subtract 100 from each side
2W=60
divide by two each side
W=30
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3 years ago

Scarlett made 1/2 of a recipe and used 4/5 cups of bananas. How many cups of bananas are required for a whole recipe??

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I think that it is 0.4

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<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B4%7D%20-%20%20%5Cfrac%7B3%7D%7B8%7Dy%20%3D%203%20%5C%5C%20" id="TexFormula

MariettaO [177]

Answer:

(6, - 4 )

Step-by-step explanation:

Given the 2 equations

$\frac{x}{4}$ - $\frac{3}{8}$ y = 3 → (1)

$\frac{5}{3}$ x - $\frac{y}{2}$ = 12 → (2)

Multiply (1) by 8 and (2) by 6 to clear the fractions

2x - 3y = 24 → (3)

10x - 3y = 72 → (4)

Rearrange (3) expressing - 3y in terms of x by subtracting 2x from both sides

- 3y = 24 - 2x

Substitute 3y = 24 - 2x into (4)

10x + 24 - 2x = 72, that is

8x + 24 = 72 ( subtract 24 from both sides )

8x = 48 ( divide both sides by 8 )

x = 6

Substitute x = 6 in either (3) or (4) and solve for y

Substituting in (3)

2(6) - 3y = 24

12 - 3y = 24 ( subtract 12 from both sides )

- 3y = 12 ( divide both sides by - 3 )

y = - 4

Solution is (6, - 4 )

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3 years ago

25% of American households have only dogs (one or more dogs) 15% of American households have only cats (one or more cats) 10% of

sergeinik [125]

Answer:

a) P=0.2503

b) P=0.2759

c) P=0.3874

d) P=0.2051

Step-by-step explanation:

We have this information:

25% of American households have only dogs (one or more dogs)

15% of American households have only cats (one or more cats)

10% of American households have dogs and cats (one or more of each)

50% of American households do not have any dogs or cats.

The sample is n=10

a) Probability that exactly 3 have only dogs (p=0.25)

$P(x=3)=\binom{10}{3}0.25^30.75^7=120*0.01563*0.13348=0.25028$

b) Probability that exactly 2 has only cats (p=0.15)

$P(x=2)=\binom{10}{2}0.15^20.85^8=45*0.0225*0.27249=0.2759$

c) Probability that exactly 1 has cats and dogs (p=0.1)

$P(x=1)=\binom{10}{1}0.10^10.90^0=10*0.1*0.38742=0.38742$

d) Probability that exactly 4 has neither cats or dogs (p=0.5)

$P(x=4)=\binom{10}{4}0.50^40.50^6=210*0.0625*0.01563=0.20508$

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3 years ago