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slamgirl [31]
3 years ago
7

.

Mathematics
1 answer:
Klio2033 [76]3 years ago
6 0

2 out of 4 statements supported the data above. This are:

JP Morgan Chase Tower has more height per story than the Key Tower.

-          This is true. Comparing the height in ft, JP Morgan Chase has a height of 1,002 ft. It is higher than the Key tower which is 947 ft only.

The Key Tower has the least height per story.

-          Looking at its height, this is the least. It only stands at 947 ft as compared to others that are more than 1,000 ft 

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Your sister is getting married this summer and she has asked you to help her select a D.J.
ioda
So Dave charges 150 at first and then 10.50 per each hour, so if your sister has the party going for say "h" hours, then his total charges will then be 150 + 10.5h.

now, Marvin's Mobile Music charges 135 at first, but then is more expensive per hour is 15 bucks per each hour, so if she has a party for "h" hours, then his total charges are 135 + 15h.

now, let's check when they're both equal, at how many hours will they cost your sister the same amount.

\bf 150+10.50h=135+15h\implies 150-135=15h-10.5h
\\\\\\
15=4.5h\implies 15=4\frac{1}{2}h\implies 15=\cfrac{9}{2}h\implies \cfrac{15}{\frac{9}{2}}=h\implies \cfrac{\frac{15}{1}}{\frac{9}{2}}=h
\\\\\\
\cfrac{15}{1}\cdot \cfrac{2}{9}=h\implies \cfrac{30}{9}=h\implies \cfrac{10}{3}=h\implies 3\frac{1}{3}=h

so, 3 and a third of a hour, or 3 hours and 20 minutes, they both will cost your sister the same amount, so it wouldn't matter who's hired if the party is only 3 hours and 20 minutes long.

however, after that, the one with higher hourly rate, Marvin's, gets more expensive, and she's better off hiring the one with cheaper hourly rate.

now, you can just get some mp3s and some loud speakers and nevermind the  DJs and you do it, or she can do it, I have the macarena song if you need that one.
3 0
3 years ago
A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

P (4 | green dice) = \dfrac{3}{6} =\dfrac{1}{2}

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

= 6 \times ( \dfrac{1}{6})^4

= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

8 0
3 years ago
Write an equation that is perpendicular to y=3/5x-1 and passes through the point (9,14)
Leokris [45]

Answer:

y=(-5/3)x+29

Step-by-step explanation:

we will use the base formula y=mx+b

In order to be perpendicular, the slope must be the flipped fraction and have the opposite sign, so our m is (-5/3)

y=(-5/3)x+b

You can plug in the (9,14) point in order to find the b

14=(-5/3)(9)+b

14=(-45/3)+b

14=(-15)+b

14+15=b

29=b

and so altogether our equation is y=(-5/3)x+29

8 0
2 years ago
Several logs are stored in a pile with 22 logs on the bottom layer, 21 on the second layer, 20 on the
Oduvanchick [21]

Answer:

[22 x 21] / 2=231 or "c"

Step-by-step explanation:

4 0
2 years ago
A) Dati cinci exemple de numere intregi mai mici ca zero.​
12345 [234]

Answer:

Numerele întregi pozitive sunt toate numerele întregi mai mari decât zero: 1, 2, 3, 4, 5, .... Numere întregi negative sunt toate opusele acestor numere întregi: -1, -2, -3, -4, -5, .... Nu considerăm că zero este un număr pozitiv sau negativ. Pentru fiecare număr întreg pozitiv, există un număr întreg negativ, iar acești numere întregi se numesc opuse. De exemplu, -3 este opusul lui 3, -21 este opusul lui 21 și 8 este opusul lui -8. Dacă un număr întreg este mai mare decât zero, spunem că semnul său este pozitiv. Dacă un număr întreg este mai mic decât zero, spunem că semnul său este negativ.

Step-by-step explanation:

4 0
2 years ago
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