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Sliva [168]
3 years ago
10

Eric prepared 45 kilograms of dough after working 9 hours. How many hours did Eric work if he prepared 55 kilograms of dough? As

sume the relationship is directly proportional
Mathematics
1 answer:
EastWind [94]3 years ago
4 0

45 / 9 = 5

55 / 5 = 11

Eric worked 11 hours to prepare 55 kilograms of dough.

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A small college has 36 math majors, 78 engineering majors , and 156 business majors in its graduate programs. What’s is the rati
Sholpan [36]

The answer is E/M+E+B or 36/270 = 4/30 = 2/15 ≈ 0.13

5 0
3 years ago
A house on the market was valued at $295,000. After several years, the value increased by 8%. By how much did the house's value
Harlamova29_29 [7]

Answer: $23,600. ; $318,600

Step-by-step explanation:

From the question, we are informed that a house on the market was valued at $295,000 and after several years, the value increased by 8%.

The value of the increase will be 8% of $295,000. This can be mathematically written as:

= 8% of $295,000

= 8/100 × $295,000

= 0.08 × $295,000

= $23,600

The current value of the house will be:

= $295,000 + $23,600

= $318,600

8 0
3 years ago
There are five seniors in a class. For each situation, write how the binomial formula is used to calculate the probability.
ololo11 [35]
A) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
<span>I choose one senior: r = 1
</span>
nCr = n! / (r! (n - r)!)
5C1 = 5! / (1! (5 - 1)!)
       = (5 * 4 * 3 * 2 * 1) / (1 * 4!)
       = 120 / (4 * 3 * 2 * 1)
       = 120 / 24
       = 5

b) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose two seniors: r = 2

nCr = n! / (r! (n - r)!)
5C2 = 5! / (2! (5 - 2)!)
       = (5 * 4 * 3 * 2 * 1) / ((2 * 1) * 3!)
       = 120 / (2 * (3 * 2 * 1))
       = 120 / (2 * 6)
       = 120 / 12
       = 10


c) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose three seniors: r = 3

nCr = n! / (r! (n - r)!)
5C3 = 5! / (3! (5 - 3)!)
       = (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * 2!)
       = 120 / (6 * (2 * 1))
       = 120 / (6 * 2)
       = 120 / 12
       = 10


d) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose four seniors: r = 4

nCr = n! / (r! (n - r)!)
5C4 = 5! / (4! (5 - 4)!)
       = (5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * 1!)
       = 120 / (24 * 1)
       = 120 / 24
       = 5


e) The answer is 1.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose five seniors: r = 5

nCr = n! / (r! (n - r)!)
5C5 = 5! / (5! (5 - 5)!)
       = (5 * 4 * 3 * 2 * 1) / ((5 * 4 * 3 * 2 * 1) * 1!)
       = 120 / (120 * 1)
       = 120 / 120
       = 1
8 0
3 years ago
Plz help!!.<br> Bahsjajajqjjajdjwk
vesna_86 [32]

Answer:

Opation b .60"

this is answers

6 0
2 years ago
4 x - 3 = -16 8.3 x + 10 = 0
Levart [38]

Answer:

idk what you asking but if it solve for for x it going be no solution

7 0
3 years ago
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