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3 years ago

* PLEASE HELP ME SOLVE THIS PROBLEM I WILL GIVE BRAINIEST*

Mathematics

1 answer:

Marta_Voda [28]3 years ago

4 0

Answer:

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In an automobile parts factory, for every 175 carburetors 3 are found to be defective. At that rate how many carbureatots out of

Furkat [3]

Answer:

At the same rate 18 carburetors out of 1,050 could be expected to be detective.

Step-by-step explanation:

The number of carburetors tested = 175

Defective pieces out o f 175 = 3

So, ratio of defected to tested = 3 : 175

Now, the number of carburetors tested in second batch = 1,050

Here, let the defective pieces = m

So, by the Ratio of Proportion,

$\frac{3}{175} = \frac{m}{1050}$

or, $m = \frac{1050 \times 3}{175} = 18$

⇒ m = 18

Hence, at the same rate 18 carburetors out of 1,050 could be expected to be detective.

8 0

3 years ago

Find the slope (-1,7),(-6,1)

natta225 [31]

Answer: m = 6/5

Step-by-step explanation:

The Formula to find m is m = y2 - y1 / x2 - x1.

So for this problem I did 7 - 1 / -1 - (-6) which would equal 6/5.

7 0

3 years ago

Someone plzzz help me i suck at math!!!<br>

Kay [80]

Answer:

B. $y=-sin\ x-2$

Step-by-step explanation:

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5 0

3 years ago

What is <img src="https://tex.z-dn.net/?f=%7Cx-6%7C" id="TexFormula1" title="|x-6|" alt="|x-6|" align="absmiddle" class="latex-f

Fofino [41]

9514 1404 393

Answer:

x - 6

Step-by-step explanation:

For values of x greater than or equal to 6, the difference x-6 is non-negative, so the absolute value function does nothing. The value is ...

x -6

7 0

3 years ago

Consider the continuous random variable x, which has a uniform distribution over the interval from 110 to 150. The probability t

Virty [35]

Answer:

The probability that x will take on a value between 120 and 125 is 0.14145

Step-by-step explanation:

For uniform distribution between a & b

Mean, xbar = (a + b)/2

Standard deviation, σ = √((b-a)²/12)

For 110 and 150,

Mean, xbar = (150 + 110)/2 = 130

Standard deviation, σ = √((150-110)²/12 = 11.55

To find the probability that x will take on a value between 120 and 125

We need to standardize 120 & 125

z = (x - xbar)/σ = (120 - 130)/11.55 = - 0.87

z = (x - xbar)/σ = (125 - 130)/11.55 = - 0.43

P(120 < x < 125) = P(-0.87 < x < -0.43)

We'll use data from the normal probability table for these probabilities

P(120 < x < 125) = P(-0.87 < x < -0.43) = P(z ≤ -0.43) - P(z ≤ -0.86) = 0.33360 - 0.19215 = 0.14145

Hope this Helps!!!

3 0

3 years ago