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3 years ago

5

A ball is thrown vertically upward from the ground with an initial velocity of 111 ft/sec. Use the quadratic function h(t) = −16

t2 + 111t + 0 to find how long it will take for the ball to reach its maximum height (in seconds), and then find the maximum height (in feet). (Round your answers to the nearest tenth.)

1 answer:

ddd [48]3 years ago

4 0

Answer:

t=3.5 seconds

Step-by-step explanation:

Given

h(t) = −16t^2 + 111t + 0

h'(t)= -32t + 111

Maximum height occurs when h(t) = 0 and the ball begins to fall

h(t)= -32t + 111=0

-32t + 111=0

-32t=-111

Divide both sides by -32

t=3.46872

Approximately, t=3.5 seconds

Recall,

Maximum height occurs when h(t) = 0

h(t)= -32t + 111=0

= -32(3.46872)+111

= -110.99904+111

= 0.00096 ft

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4 years ago

Can 9.48 be the probability of an outcome in a sample space?

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3 years ago

Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to

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Answer:

(a) The probability that X is at most 30 is 0.9726.

(b) The probability that X is less than 30 is 0.9554.

(c) The probability that X is between 15 and 25 (inclusive) is 0.7406.

Step-by-step explanation:

We are given that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked. A random sample of 200 shafts is taken.

Let X = <u><em>the number among these that are nonconforming and can be reworked</em></u>

The above situation can be represented through binomial distribution such that X ~ Binom(n = 200, p = 0.11).

Here the probability of success is 11% that this much % of all steel shafts produced by a certain process are nonconforming but can be reworked.

Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).

So, the new mean of X, $\mu$ = $n \times p$ = $200 \times 0.11$ = 22

and the new standard deviation of X, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{200 \times 0.11 \times (1-0.11)}$

= 4.42

So, X ~ Normal( $\mu =22, \sigma^{2} = 4.42^{2}$ )

(a) The probability that X is at most 30 is given by = P(X < 30.5) {using continuity correction}

P(X < 30.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{30.5-22}{4.42}$ ) = P(Z < 1.92) = <u>0.9726</u>

The above probability is calculated by looking at the value of x = 1.92 in the z table which has an area of 0.9726.

(b) The probability that X is less than 30 is given by = P(X $\leq$ 29.5) {using continuity correction}

P(X $\leq$ 29.5) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{29.5-22}{4.42}$ ) = P(Z $\leq$ 1.70) = <u>0.9554</u>

The above probability is calculated by looking at the value of x = 1.70 in the z table which has an area of 0.9554.

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