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3 years ago

(N^3 - N^4) - (3n^3- 7n^4)

1 answer:

3 0

If you would like to solve (n^3 - n^4) - (3n^3 - 7n^4), you can do this using the following steps:

(n^3 - n^4) - (3n^3 - 7n^4) = n^3 - n^4 - 3n^3 + 7n^4 = n^3 - 3n^3 - n^4 + 7n^4 = -2n^3 + 6n^4

The correct result would be -2n^3 + 6n^4.

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Abraham told his friends to try and guess his number. He said that when you take his number and multiply it by 4 and add 12 you

lisov135 [29]

Answer:

C) 22

Step-by-step explanation:

I have set up an equation based off of what Abraham said.

$(n*4)+12=100\\\\4n+12=100\\\\4n+12-12=100-12\\\\4n=88\\\\\frac{4n=88}{4}\\\\\boxed{n=22}$

The number should be 22.

Hope this helps.

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3 years ago

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What type of graph is shown, and what is the growth factor?

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linear function; growth factor of 4

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3 years ago

Are there any perfect squares that go into 380?

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I belive the answer is no. Tell me if im doing this wrong, but the square root of 380 is a decimal number.

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3 years ago

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specifica

Natalija [7]

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal( $\mu=0.753,\sigma^{2} =0.004^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = population mean = 0.753 inch

$\sigma$ = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X $\leq$ 0.75 inch)

P(X < 0.753) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.753-0.753}{0.004} } }$ ) = P(Z < 0) = 0.50

P(X $\leq$ 0.75) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.75-0.753}{0.004} } }$ ) = P(Z $\leq$ -0.75) = 1 - P(Z < 0.75)

= 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X $\leq$ 0.74 inch)

P(X < 0.75) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.75-0.753}{0.004} } }$ ) = P(Z < -0.75) = 1 - P(Z $\leq$ 0.75)

= 1 - 0.7734 = 0.2266

P(X $\leq$ 0.74) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.74-0.753}{0.004} } }$ ) = P(Z $\leq$ -3.25) = 1 - P(Z < 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

P(X > 0.76) = P( $\frac{X-\mu}{\sigma} } }$ > $\frac{0.76-0.753}{0.004} } }$ ) = P(Z > -1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

P(X < 0.74) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.74-0.753}{0.004} } }$ ) = P(Z < -3.25) = 1 - P(Z $\leq$ 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

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4 years ago

Explain how you can find 4x754 different two way

svetlana [45]

4 of 754 and 754 of 4. Also, try different strategies of showing work.

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3 years ago

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