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Lapatulllka [165]

2 years ago

9

Consider all seven-digit numbers that can be created from the digits 0-9 where the first and last digits must be odd and no digi

t can repeat. What is the probability of choosing a random number that starts with 3 from this group

1 answer:

Natali [406]2 years ago

6 0

Answer:

Considering all seven-digit numbers that can be created from the digits 0-9 I could deduce that you can make 100 , 700, 800 even infinite numbers

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You have just timed a person doing a hair cut for the first time. It took 50 minutes. What unit improvement factor learning curv

Sladkaya [172]

Answer: C. 70 percent

Step-by-step explanation:

Given, Time for the first unit = 50 minutes

Time for the second unit = 35 minutes

The unit improvement factor learning curve = (The time for the second unit) ÷ (time for the first unit) x 100.

So, The unit improvement factor learning curve = 35÷ 50 × 100 = 70 percent.

Hence, the correct option is "C. 70 percent".

3 0

2 years ago

A local pizzeria offers 15 toppings for their pizzas and you can choose any 3 of them for one fixed price. How many different ty

Shtirlitz [24]

Answer:

455 or 680, depending

Step-by-step explanation:

If we assume the three choices are different, then there are ...

15C3 = 15·14·13/(3·2·1) = 35·13 = 455

ways to make the pizza.

___

If two or three of the topping choices can be the same, then there are an additional ...

2(15C2) +15C1 = 2·105 +15 = 225

ways to make the pizza, for a total of ...

455 + 225 = 680

different types of pizza.

__

There is a factor of 2 attached to the number of choices of 2 toppings, because you can have double anchovies and tomato, or double tomato and anchovies, for example, when your choice of two toppings is anchovies and tomato.

_____

nCk = n!/(k!(n-k)!)

6 0

2 years ago

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

2 years ago

joe is responsible for reserving hotel rooms for a company trip. His company changes plans and increases how many people are goi

Andreyy89

Answer:

Joe reserves 5 more blocks.

Step-by-step explanation:

50-16=34.

34/8=4.25

You must round up 4.25 to 5 and 16 rooms equal 2 blocks.

7 0

2 years ago

Read 2 more answers

Triangle N T M is shown. Angle N T M is a right angle. An altitude is drawn from point T to point U on side N M to form a right

stich3 [128]

Answer:

The value of y is 6 $\sqrt{3}$ units ⇒ 2nd answer

Step-by-step explanation:

In the attached figure

∵ ∠MTN is a right angle

∵ TU is the altitude of the triangle

- There are some rules in this triangle let us revise them

(NT)² = NU . NM
(MT)² = MU . MN
(TU)² = MU . NU
TM . TN = TU . MN

∵ NU = 9 units

∵ UM = 3 units

∵ MN = UM + NU

∴ MN = 3 + 9 = 12 units

- By using the 1st rule above

∴ (NT)² = 9 × 12

∴ (NT)² = 108

- Take a square root to both sides

∴ NT = $\sqrt{108}$

- Simplify the root

∴ NT = 6 $\sqrt{3}$ units

∵ NT is y

∴ y = 6 $\sqrt{3}$ units

The value of y is 6 $\sqrt{3}$ units

8 0

2 years ago

Read 2 more answers