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2 years ago

All 136 students in the Math Club went on a field trip. Some students rode in vans which hold 8

1 answer:

5 0

Answer: 3 buses, 2 vans
Explanation: let x be number of buses, y be number of vans,
136= 40x+8y
5=x+y
Use simultaneous equation,
136-8(5)= 40x +8y - 8(x+y)
96= 32x
x = 3
When x=3, y= 2

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June has $1.95 in dimes and nickels. She has a total of 28 coins. how many of each type of coin does she have?

Dovator [93]

Answer:

11 dimes, 17 nickels

Step-by-step explanation:

d + n = 28, put one variable by itself on a side, d = 28 - n

.10d + .05n = 1.95

Sub the first equation into the second

.10(28 - n) + .05n = 1.95

2.8 - .10n + .05n = 1.95

2.8 - .05n = 1.95

-.05n = -0.85

n = 17 nickels

d + 17 = 28

d = 11 11 dimes

7 0

2 years ago

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Can you help me, if you get it right, you get brainlest

SVEN [57.7K]

Can you be more specific? I need a question and I need a close up ! Lol but I’m trying to help.

4 0

2 years ago

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第 5 个问题 a multiple choice exam has 10 questions. each question has 3 possible answers, of which one is correct. a student knows

tatuchka [14]

The chances that the student was merely guessing is 1/3.

Bayes Theorem determines the conditional probability of an event A given that event B has already occurred.

denoted by

$P(A/B)=\frac{P(A)*P(B/A)}{P(B)}$

let A be the event that the student knows the answer .

B be the event that the student does not knows the answer .

and

E be the event he gets answer correct .

According to the given question

$P(A)=\frac{4}{10} \\\\ P(B)=1-\frac{4}{10} =\frac{6}{10}$

Probability that the answer is correct ,given that he knows the answer is

$P(E/A)=1$

Probability that the answer is correct ,given that he guesses it is

$P(E/B)=\frac{1}{3}$ [as the MCQ has 3 options and only one is correct]

We need to find the probability that he guesses the answer given that it is correct.

Required probability $P(B/E)=\frac{P(B)*P(E/B)}{P(A)*P(E/A)+P(B)*P(E/B)}$

Substituting the values we get

$P(B/E)=\frac{\frac{6}{10} *\frac{1}{3} }{\frac{4}{10} *1+\frac{6}{10} *\frac{1}{3} }$

$=\frac{6}{30}*\frac{30}{18} \\ \\ =\frac{6}{18} \\ \\ =\frac{1}{3}$

Therefore , the chances that the student was merely guessing is 1/3.

Learn more about Probability here brainly.com/question/13140147

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8 0

9 months ago

Robin has triple as many marbles Amy. Sam has 15 more marbles than robin. Together they all have 99 marbles. How many marbles do

tigry1 [53]

Let: x be Amy's marbles
3x be Robin's marbles
15+3x be Sam's marbles
x+3x+15+3x=99
7x+15=99
7x=99-15 (transfer 15 to the right side of the equation, then minus)
7x/7=84/7 (divide both sides of the equation by 7)
x=12
Now substitute:
x = 12 (Amy's)
3x = 3*12 = 36 (Robin's)
15+3x = 15+3*12 = 51 (Sam's)
12+36+51 = 99
Therefore, Amy has 12 marbles, Robin has 36 marbles and Sam has 51 marbles.

5 0

3 years ago

Write the sum of 83 and<br> a numbers in a mathematical <br> atical way.

vaieri [72.5K]

Answer:

$83 + n$

Step-by-step explanation:

The sum of 83 and a number would be $83 + n$ as an expression, where $n$ is "said number".

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3 years ago

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