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Pie
3 years ago
8

At a local store, bananas cost 0.54 per pound , apples cost 2.19 per pound , and oranges cost 1.55 per pound . Aaron bought 3 po

unds of each how much money did Aaron pay for his fruit
Mathematics
2 answers:
Alex777 [14]3 years ago
8 0

$0.54 x 2.6 =$1.404  

$1.28 x 3.1=$3.968  

$1.404 rounds to $1.40

$3.968 rounds to $4

$4 = $ 1.40 = $5.40  

:)

Orlov [11]3 years ago
4 0

Answer:

$12.84

Step-by-step explanation:

to solve this you will just have to multiply the given prices by the three pounds and then add up all the fruits:

0.54 x 3 = 1.62

2.19 x 3= 6.57

1.55 x 3= 4.65

1.62 + 6.57 + 4.65 = 12.84

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oksano4ka [1.4K]
Divide: 9x 2/36 - 4y 2/36 = 36/36. Simplify: x2/4 - y2/9 = 1. Then do (x-0) 2/4 - (y-0)2 =1. Hyperbola.
4 0
3 years ago
Nancy is making lemonade for her birthday party. She made 6 batches. Each batch of lemonade uses 24 fluid ounces of lemon juice
Goryan [66]

Answer:

Step-by-step explanation:

1 Us pint = 16 Us fluid ounces

It means that 8 pints of water would be

8 × 16 = 128 fluid ounces

Each batch of lemonade uses 24 fluid ounces of lemon juice and 8 pints of water. The total number of fluid ounces in each batch is

128 + 24 = 152 fluid ounces

Also,

1 fluid ounce = 0.125 Us cups

Therefore, the number of cups made is

152 × 0.125 = 19 cups

If each serving is 1 cup, then the number of servings that she made from each batch of lemonade is 19

servings. The number of batches made is 6. Therefore, the number of servings made altogether is

6 × 19 = 114

If her brother came and spilled half of it on the floor, then the number of servings that she have left after that for her party is

114/2 = 57 servings

7 0
3 years ago
Please help me find x<br> Need help fast!!
Bond [772]

Answer:

x = 100°

Step-by-step explanation:

p and q are parallel lines. Construct line 'l' parallel to p and q

a = 30°   {p// l  when parallel lines are intersected by transversal alternate interior angles are congruent}

c +110 = 180  {linear pair}

c = 180  - 110

c = 70°

b= c   {q // l  when parallel lines are intersected by transversal alternate interior angles are congruent}

b = 70°

x = a + b

x = 30° + 70°

x = 100°

6 0
3 years ago
Flying fish use their pectoral fins like airplane wings to glide through the air. Suppose a flying fish reaches a maximum height
GrogVix [38]

The flight is in the shape of a parabola with a vertex 5 feet above the water and  1/2 * 33 = 16.5 feet horizontally from the point of leaving the water

y = a(x - h)^2 + k

where  (h,k)  is the vertex of the  parabola and here it is (5 , 16.5), so we have the function:-

y = a(x - 16.5)^2 + 5

when x = 0 y = 0  so

0 = a(-16.5)^2 + 5

which gives a = -0.018365

So our function for the flight path is

y = -0.018365(x - 16.5)^2 + 5     Answer



7 0
3 years ago
Read 2 more answers
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
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