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3 years ago

At a pizza shop, 80% of the customers order a pizza, 15% of the customers

Mathematics

2 answers:

meriva3 years ago

5 0

90% out of 105% which would be 0.85

Free_Kalibri [48]3 years ago

5 0

Answer: 0.85

Step-by-step explanation:

Let A be the event that the customer orders a pizza and B be the event that the customer orders a salad.

Then , we have $P(A)=80\%=0.80$

$P(B)=15\%=0.15$

$P(A\cap B)=10\%=0.10$

Now, the probability that he or she orders either a pizza or a salad is given by :-

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\\Rightarrow\ P(A\cup B)=0.80+0.15-0.10=0.85$

Hence, the required probability = 0.85

You might be interested in

$600is what percent of $1000

GrogVix [38]

Answer:

60%

Step-by-step explanation:

$600 as a percentage of $1000

= (600/1000) x 100%

= 60%

Hope this helps!

5 0

3 years ago

Read 2 more answers

Four lawn sprinkler heads are fed by a 1.9-cm-diameter pipe. The water comes out of the heads at an angle of 35° above the horiz

klemol [59]

Answer:

Step-by-step explanation:

Given:

Angle, θ = 35°

Vertical distance, Δx = 6 m

Diameter, d = 1.9 cm

= 0.019 m

When the water leaves the sprinkler, it does so at a projectile motion.

Therefore,

Using equation of motion,

(t × Vox) = 2Vo²(sin θ × cos θ)/g

= Δx = 2Vo²(sin 35 × cos 35)/g

Vo² = (6 × 9.8)/(2 × sin 35 × cos 35)

= 62.57

Vo = 7.91 m/s

Area of sprinkler, As = πD²/4

Diameter, D = 3 × 10^-3 m

As = π × (3 × 10^-3)²/4

= 7.069 × 10^-6 m²

V_ = volume rate of the sprinkler

= area, As × velocity, Vo

= (7.069 × 10^-6) × 7.91

= 5.59 × 10^-5 m³/s

Remember,

1 m³ = 1000 liters

= 5.59 × 10^-5 m³/s × 1000 liters/1 m³

= 5.59 × 10^-2 liters/s

= 0.0559 liters/s.

For the 4 sprinklers,

The rate at which volume is flowing in the 4 sprinklers = 4 × 0.0559

= 0.224 liters/s

Area of 1.9 cm pipe, Ap = πD²/4

= π × (0.019)²/4

= 2.84 × 10^-4 m²

Volumetric flowrate of the four sprinklers = 4 × 5.59 × 10^-5 m³/s

= 2.24 × 10^-4 m³/s

Velocity of the water, Vw = volumetric flowrate/area

= 2.24 × 10^-4/2.84 × 10^-4

= 0.787 m/s

7 0

3 years ago

Read 2 more answers

The gravitational force exerted by the planet Earth on a unit mass at a distance r from the center of the planet is:

jolli1 [7]

Answer:

Yes, F is a continuous function of r

Step-by-step explanation:

We are given that

When r<R

$F(r)=\frac{GMr}{R^3}$

$r\geq R$

$F(r)=\frac{GM}{r^2}$

Where M=Mass of the earth

R=Radius of earth

G=Gravitational constant

We have to find the function is continuous of r or not.

LHL

$\lim_{r\rightarrow R-}\frac{GMr}{R^3}=\frac{GMR}{R^3}=\frac{GM}{R^2}$

RHL

$\lim_{r\rightarrow R+}\frac{GM}{R^2}=\frac{GM}{R^2}$

$F(R)=\frac{GM}{R^2}$

When a function is continuous at x=a

Then, LHL=RHL=f(a)

RHL==LHL=F(R)

Hence, the function is continuous of r.

8 0

3 years ago

If function F is defined by f(x) = 3x^2 + 5, what is f(x-4)

shepuryov [24]

Answer:

The function f(x) when x=x-4 is $f(x-4)=3x^2-24x+53$ or $f(x-4)=3x(x-8)+53$

Step-by-step explanation:

Given function f is defined by $f(x)=3x^2+5$

Now to find the function f(x-4):

$f(x)=3x^2+5$

Put x=x-4 in the above function we get

$f(x-4)=3(x-4)^2+5$

$f(x-4)=3(x^2-2(x)(4)+4^2)+5$ (using the formula $(a-b)^2=a^2-2ab+b^2$ )

$f(x-4)=3(x^2-8x+16)+5$

$f(x-4)=3x^2-24x+48+5$ (adding the like terms)

$f(x-4)=3x^2-24x+53$

$f(x-4)=3x(x-8)+53$

Therefore $f(x-4)=3x^2-24x+53$ or $f(x-4)=3x(x-8)+53$

The function f(x) when x=x-4 is $f(x-4)=3x^2-24x+53$ or $f(x-4)=3x(x-8)+53$

4 0

3 years ago

How dose this work out

Morgarella [4.7K]

Can you take a better picture. That would help to answer the question.

7 0

3 years ago