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3 years ago

A

Mathematics

1 answer:

VLD [36.1K]3 years ago

4 0

it is more than unlikely this will be answered. your format made it way to hard to read

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Does the table represent a function?(Only 2 hours to answer!)

alisha [4.7K]

Answer: (a) Yes. The table is a function.

Step-by-step explanation:

In order to be a function, there cannot be any duplicate x-values.

In the given table, each x-value is different so this IS a function.

6 0

3 years ago

At the county fair, Darius can purchase 3 candy apples and 4 bags of peanuts for $11.33. He can purchase 9 candy apples and 5 ba

Ksenya-84 [330]

Let c represents the cost of a candy apple and b represents the cost of a bag of peanuts.

Darius can purchase 3 candy apples and 4 bags of peanuts. So his total cost would be 3c + 4b. Darius can buy 3 candy apples and 4 bags of peanuts in $11.33,so we can write the equation as:

3c + 4b = 11.33 (1)

Darius can purchase 9 candy apples and 5 bags of peanuts. So his total cost would be 9c + 5b. Darius can buy 9 candy apples and 5 bags of peanuts in $23.56,so we can write the equation as:

9c + 5b = 23.56 (2)

Darius decides to purchase 2 candy apples and 3 bags of peanuts. The total cost in this case will be 2c + 3b. To find this first we need to find the cost of each candy apple and bag of peanuts by solving the above two equations.

Multiplying equation 1 by three and subtracting equation 2 from it, we get:

3(3c + 4b) - (9c + 5b) = 3(11.33) - 23.56

9c + 12b - 9c - 5b = 10.43

7b = 10.43

b = $1.49

Using the value of b in equation 1, we get:

3c + 4(1.49) = 11.33

3c = 5.37

c = $ 1.79

Thus, cost of one candy apple is $1.79 and cost of one bag of peanuts is $1.49.

So, 2c + 3b = 2(1.79) + 3(1.49) = $ 8.05

Therefore, Darius can buy 2 candy apples and 3 bags of peanuts in $8.05

3 0

3 years ago

a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade

son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

$P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})$

$P(at least 1 dull blade)=1-0.2931$

P(at least 1 dull blade)=0.7068

5 0

3 years ago

An expression is shown below. (x^3/4)(x^2/3)

Mkey [24]

For all exponents, (a^n a^m) = a^(n+m)
apply same rules (X^3X^2)/(4*3)
Combine powers. (X^3X^2)/4*3 = X^(3+2)/4*3
X^5/12

3 0

3 years ago

Sin167cos107-cos167sin107

insens350 [35]

Sin 167 = 0.22495

cos 107 = -0.29237

cos 167 = -0.97437

sin 107 = 0.95630

(sin 167)(cos 107) - (cos 167)(sin 107)

(0.22495)(-0.29237) - ( -0.97437)(0.95630)

= 0.86602

hope this helps :)

6 0

3 years ago