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RideAnS
14 days ago
9

Grades 3, 4, and 5 have their annual field day together. Each grade level is given 16 gallons of water.

Mathematics
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If you ask three strangers about their birthdays, what is the probability:
Inessa [12570]
Part A:

The probability that all three strangers have their birthdays on a Wednesday is calculated as

\left( \frac{1}{7} \right)^3= \bold{\frac{1}{343}}



Part B:

The probability that the birthdays of the three individuals fall on distinct days throughout the week is calculated as

\left( \frac{1}{7} \right)\left( \frac{1}{6} \right)\left( \frac{1}{5} \right)= \bold{\frac{1}{210}}



Part C:

The probability that none of the three have their birthdays on a Saturday is determined by

\left( \frac{6}{7} \right)^3= \bold{\frac{216}{343}}
8 0
1 month ago
Isabella built a time travel machine, but she can't control the destination of her trip. Each time she uses the machine she has
babunello [11817]

The likelihood that at least one trip occurs before Isabella's birth is 0.7627.

Step-by-step explanation:

In this scenario, Isabella has invented a time machine, but she lacks control over where she travels. Each use of the device holds a 0.25 probability of leading her to a time preceding her birth. Over the initial year of trials, she operates her machine 5 times. If we assume every journey has an equal chance of going back in time, we can calculate the odds that at least one of these trips occurs before she was born. Here's the calculation:

The probability of traveling to a time prior to her birth is 0.25.

The chance of not traveling back in time, given that the machine is used 5 times:

⇒ (1-0.25)(1-0.25)(1-0.25)(1-0.25)(1-0.25)

⇒ (1-0.25)^5

⇒ (0.75)^5

The probability that at least one trip goes before Isabella's birth is equal to 1 minus the probability of not traveling back to that period:

⇒ 1-(0.75)^5

⇒ 0.7627

Consequently, the chance that at least one trip travels before Isabella's birth is 0.7627.

4 0
2 months ago
Read 2 more answers
Match each step to its justification to solve 2x+5=19
Inessa [12570]
It includes the procedures to reach that specific point.
Some of them are arranged incorrectly.
I will label them with letters.

The sequence is:
A: 2x+5=19
B: 2x+5-5=19-5
C: 2x=14
D: 2x/2=14/2
E: x=7


A: initial condition
B: we subtracted 5 from both sides based on the subtraction property of equality
C: perform the subtraction
D: applying the division property of equality (dividing both sides by 2)
E: result of the division
5 0
2 months ago
Read 2 more answers
Solve the following addition and subtraction problems. a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam b. 5 sq.km 95 ha 8,994 sq.m + 11 sq.
Svet_ta [12734]
To tackle the problem, the general approach is to convert all measurements into the smallest unit possible.
a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam
3,000 m 900 m 90 m 19 m + 7,000 m 70 m = 4,009 + 7,070 = 11,079 m
b. 5 sq.km 95 ha 8,994 sq.m + 11 sq. km. 11 ha 9,010 sq. m.
5,000,000 sq m 95,0000 sq m 8,994 sq m + 11,000,000 sq m 110,000 sq
9,010 sq m
5,103,994 sq m + 11,119,010 sq m = 16,223,004 sq m
c. 44 m - 5 dm
44 m - 0.5 m = 43.5 m
d. 73 km 47 hm 2 dam - 11 km 55 hm

73,000 m 4,700 m 20 m - 11,000 m 5,500 m
77,720 m - 16,500 m = 61,220 m
5 0
1 month ago
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