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3 years ago

Helllllpppp meeeee I neeedddd hellllpppp nowwww pleaseeee

Mathematics

1 answer:

Crank3 years ago

8 0

Answer:

Step-by-step explanation:

They are all integers so they are rational numbers, they are whole numbers as well except there are negative numbers so they are not all natural.

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3.4 mi/h converted into ft/sec<br><br> Please someone help me out on this question!

Vlad [161]

To convert mi/h to ft/sec, multiply it to 1.466.

So 3.4 mi/h x 1.466 = 4.986 ft/sec

5 0

3 years ago

Find X and Y <br> I need help plz ASAP

Nezavi [6.7K]

Answer:

x=5.657 y=5.657

Step-by-step explanation:

4 0

3 years ago

Beth poured the same amount of juice into 10 glasses.She used 2 quarts of juice.How much juice is in each glass

agasfer [191]

6.4 ounces.

You can get this by first converting to ounces which is a smaller value. Then you can divide by 10.

7 0

3 years ago

Each vertical cross-section of the triangular

Blizzard [7]

Answer:

The vertical height is $h=3\ units$

Step-by-step explanation:

we know that

In the right triangle ABC

see the attached figure

To find the height h apply the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

we have

$c=AB=3\sqrt{2}\ units$

$a=BC=6/2=3\ units$

$b=AC=h\ units$

substitute

$(3\sqrt{2})^{2}=3^{2}+h^{2}$

$18=9+h^{2}$

$h^{2}=18-9=9$

$h=3\ units$

3 0

3 years ago

Read 2 more answers

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien

diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = $\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}$

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) = $\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}$ = $\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}$ = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0

3 years ago