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stepladder [879]
3 years ago
5

I need help......................

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

a) false

b) true

c) true

d) false

hope dis helps ^-^

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A container contains 12 diesel engines. The company chooses 5 engines at​ random, and will not ship the container if any of the
Rufina [12.5K]

Answer:\frac{^{10}C_5}{^{12}C_5}

Step-by-step explanation:

Given a total of 12 diesel engines

out of which 2 are defective

Company have to choose 5 good engines to ship.

Therefore no of ways in good engines are selected is ^{10}C_5

no of ways in which any 5 engines are selected from a total of 12 engines is

^{12}C_5

Therefore the required probability =\frac{^{10}C_5}{^{12}C_5}

=\frac{7}{22}=0.318

3 0
3 years ago
Who can help me with number 14
Rzqust [24]
(Original Price $425.4)

$328 × 30% = 98.4
328 + 98.4 = $426.40
5 0
4 years ago
Read 2 more answers
For the animal shelter, Elena Nppurchased
Naddik [55]

Elena's total cost (C) is 5(9.98 + 24.99 + x)

Since each dog leashes cost $9.98, each bag of dog food cost $24.99 and each crate of egg cost $x, hence:

Total cost (C) = 5 * $9.98 + (5 * 24.99) + (5 * x)

C = 5(9.98 + 24.99 + x)

Therefore Elena's total cost (C) is 5(9.98 + 24.99 + x)

Find out more at: brainly.com/question/21105092

6 0
3 years ago
The area of the sector of a circle with a radius of 8 centimeters is 125.6 square centimeters. The estimated value of Pi is 3.14
elena55 [62]
Hello there!

The correct answer is C. 225.

Hope This Helps You!
Good Luck :)
6 0
3 years ago
Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati
Brrunno [24]
Remember that <span>an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.
 
Lets solve our equation to find out what is the extraneous solution:
</span>\sqrt{x-3} =x-5
(\sqrt{x-3})^2 =(x-5)^2
x-3=x^2-10x+25
x^2-11x+28=0
(x-4)(x-7)=0
x-4=0 and x-7=0
x=4 and x=7
<span>
So, the solutions of our equation are </span>x=4 and x=7. Lets replace each solution in our original equation to check if they are valid solutions:
- For x=7
\sqrt{x-3} =x-5
\sqrt{7-3} =7-5
\sqrt{4} =2
2=2
We can conclude that 7 is a valid solution of the equation.

- For x=4
\sqrt{x-3} =x-5
\sqrt{4-3} =4-5
\sqrt{1} =1
1 \neq 1
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: <span>D. the extraneous solution is x = 4</span>
7 0
3 years ago
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