Please help me with this!!!!!

Hera spent $16 on dinner then she spent $6 on dessert and some more on a taxi ride including the taxi fair she spent a total of

s344n2d4d5 [400]

She spent $9.

$16+$6 is $22, so the total $31 minus $22 leaves us with the $9 she spent for the taxi.

8 0

3 years ago

Read 2 more answers

Bill has a perfect cube of an unknown substance. The density of the substance is 3.0 g/cm. He

Marina CMI [18]

Answer:

9cm

Step-by-step explanation:

v=m/d

81/3=27cm^3

vol. of cube is h×a×b

since it is a cube all sides are the same

27=3h

h=9cm

6 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Len [333]

Answer: 3

Step-by-step explanation:

Given : Standard deviation : $\sigma=4.1\text{ seats}$

Margin of error : $E=\pm5\text{ seats}$

Significance level : $\alpha: 1-0.95=0.05$

By using the standard normal table of z ,

Critical value : $z_{\alpha/2}=1.96$

The formula we use to find the minimum sample size required :-

$n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2$

i.e. $n=(\dfrac{(1.96)(4.1)}{5})^2=2.58309184\approx3$

Hence, the number of lights should we select if we wish to estimate μ to within 5 seats and be 95 percent confident =3

6 0

3 years ago

There is a small island in the middle of 100m wide river and a tall tree stands on the island. P and q are points directly oppos

Savatey [412]

Answer:

90

Step-by-step explanation:

8 0

3 years ago

About A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows: A: The sum on the two

Flauer [41]

$p(A) = \frac{1}{2}$

$p(B) = \frac{1}{6}$

$p(C) = \frac{1}{6}$

$P(A | C)=\frac{1}{2}$

Solution:

The probability of an event is given as:

$\text { probability of an event }=\frac{\text { number of favorable outcomes }}{\text { total number of outcomes }}$

In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}

First let us calculate p(A):

The event is defined as: The sum on the two dice is even

Sum on two dice is even if and only if either both dice turn up odd or both even.

The odd outcomes in thowing a single die = 3 {1, 3, 5}

The even outcomes in throwing a single die = 3 {2, 4, 6}

The probability that both turn up odd is:

$\text { probability of both die turing up odd }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

Similarly, the probability that both turn up even is:

$\text { probability of both die turing up even }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

probability that the sum on two dice is even = probability that both turn up odd + probability that both turn up even

$\text { probability of sum on two dice is even }=\mathrm{p}(\mathrm{A})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Thus $p(A) = \frac{1}{2}$

Let us calculate p(B):

The event B is defined as: The sum on the two dice is at least 10

The total possible outcomes of two die is given as:

{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :

atleast 10 means that sum can be 10 or greater than 10

{(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)}

Here favourable outcomes = 6

Total number of outcomes = 36

Hence, the probability that the sum of the two dice will be at least 10 is:

$\text { probability that the sum of the two dice will be at least } 10=\frac{6}{36}=\frac{1}{6}$

Thus $p(B) = \frac{1}{6}$

Let us calculate p(C):

The event C is defined as: The red die comes up 5

Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

$\text { probability of red die comes up } 5 \text { is the event }=\frac{6}{36}=\frac{1}{6}$

Thus $p(C) = \frac{1}{6}$

B) What is p(A l C)

$P(A | C)=\frac{p(A \cap C)}{P(C)}$

$\mathrm{A} \cap \mathrm{C}=\{(1,5),(3,5),(5,5)\}$

$p(A \cap C)=\frac{3}{36}=\frac{1}{12}$

$P(A | C)=\frac{p(A \cap C)}{P(C)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}$

Thus $P(A | C)=\frac{1}{2}$

8 0

4 years ago