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slava [35]
3 years ago
10

Help ASAP I’ll mark as brainlister

Mathematics
2 answers:
zavuch27 [327]3 years ago
4 0

Answer:

its 10 i think

eimsori [14]3 years ago
3 0

Answer:

10

Step-by-step explanation:

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What is the value of y
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Answer:

y= -6

Step-by-step explanation:

-2y - 5y= -7y

-7y=42

42/-7=

y= -6

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2 years ago
The data set of water consumption for a small town in gallons per day is listed below. What is the effect on the mean and standa
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Find the area and circumference of a circle with a radius of 4 cm. Use the value 3.14 for pi.
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C= 25.12 cm

Step-by-step explanation:

7 0
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 = <u><em>diameter of the ball bearings</em></u>

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 = <u>0.2734</u>.

(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 = <u>0.226</u>.

(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 = <u>0.0401</u>

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 = <u>0.0006</u>

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.

8 0
3 years ago
Need help with #5.<br> Find angle and solve for x
Leona [35]

Answer:

see explanation

Step-by-step explanation:

The 2 marked angles are vertical and congruent, thus

24x = 23x + 5 ( subtract 23x from both sides )

x = 5

Thus angles = 24 × 5 = 120°

6 0
3 years ago
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