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Nataly [62]
3 years ago
10

The equation h(t)=−16t2+120t+65 gives the height of a firework, in feet, t seconds after it is launched from a platform.

Mathematics
2 answers:
kakasveta [241]3 years ago
8 0

Answer:

The height of the platform is 65 ft

Step-by-step explanation:

Let

t -----> the time in seconds after the launch of the fireworks from a platform

h(t) ----> the height of a firework in feet

we have

h(t)=-16t^{2}+120t+65

we know that

The height of the platform is when the value of t is equal to zero (y-intercept)

For t=0

h(0)=-16(0)^{2}+120(0)t+65

h(0)=65\ ft

therefore

The height of the platform is 65 ft

Ira Lisetskai [31]3 years ago
8 0

Answer:

65 ft

Step-by-step explanation:

We are given that an equation which gives the height of firework

h(t)=-16t^2+120t+65

Where h(t)=Height in feet

t=Time in seconds

Substitute t= 0

h(0)=65 ft

When t= 0 then height of platform =h(0)

Hence, the height of platform=65 ft

Answer: 65 ft

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

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\frac{12}{64} =\frac{x}{800}

Cross multiply:

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