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3 years ago

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 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:
 $\sqrt{x-3} =x-5$
$(\sqrt{x-3})^2 =(x-5)^2$
$x-3=x^2-10x+25$
$x^2-11x+28=0$
$(x-4)(x-7)=0$
$x-4=0$ and $x-7=0$
$x=4$ and $x=7$

So, the solutions of our equation are $x=4$ 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: D. the extraneous solution is x = 4

7 0

3 years ago