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3 years ago

The cost of 10 oranges is $1.00. What is the cost of 5 dozen oranges?

2 answers:

5 0

They said 5 dozen. So 1 dozen is 12 12*5= 60 10 orange is $1 20 orange is $2 30 orange is $3 40 orange is $4 50 orange is $5 60 orange is $6 ( So the answer is $6) Hope it's help you easy way

Crank3 years ago

3 0

12 x 5 = 60 so The cost of 60 oranges is $6.00.

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AB = 6cm , AC = 12cm <br> Calculate the length of CD <br> Give your answer to 3 significant figures

Len [333]

Answer:

144-36=108

Square root 108 = 10.4cm (cb)

Soh Cah toa

Sin(55) =o/h

=10.4/h

H x sin55 =10.4

÷sin55

H= 10.4/sin55

H=12.7cm

CD is 12.7cm to 3 s.f.

6 0

2 years ago

How do you solve this ?

Anettt [7]

Answer:

Nice handwriting :)

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Step-by-step explanation:

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3 years ago

Read 2 more answers

F(x) = 3x² + 5x - 2<br> g(x) = 5x³-4x² + 4<br> Find (f + g)(x)

natali 33 [55]

Answer: $5x^3 - x^2 + 5x+2$

Step-by-step explanation:

$(f+g)(x)=(3x^2 + 5x-2)+(5x^3 - 4x^2 +4)=3x^2 + 5x-2+5x^3 - 4x^2 +4=\boxed{5x^3 - x^2 + 5x+2}$

7 0

2 years ago

If you have 10 red, 10 blue, 10 orange and 10 black socks in a drawer, how many socks must you pull out of the drawer (assume it

Orlov [11]

Answer:

Step-by-step explanation:

The probability of u choosing two orange socks from the sock drawer WITHOUT REPLACEMENT:

= P(the first one is Orange) * P(the second one is Orange)

= 10/40 * 9/39 = 3/52.

6 0

3 years ago

) Suppose that a subset of five balls will be randomly selected from an urn containing amber, blue, and green balls. (a) If the

vazorg [7]

Answer:

0.0023 = 0.23% probability that all 5 balls selected will be the same color

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the balls are selected is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes:

Either 5 amber from a set of 6, or 5 blue from a set of 5. So

$D = C_{6,5} + C_{5,5} = \frac{6!}{5!1!} + \frac{5!}{5!0!} = 6 + 1 = 7$

Total outcomes:

5 balls selected from a set of 6 + 5 + 4 = 15. So

$T = C_{15,5} = \frac{15!}{5!10!} = 3003$

Probability:

$p = \frac{D}{T} = \frac{7}{3003} = 0.0023$

0.0023 = 0.23% probability that all 5 balls selected will be the same color

5 0

3 years ago