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NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.

Aleks [24]

Answer:

A.) 24.08 seconds

B.) 825.42 metres

Step-by-step explanation:

function of time is given as

h ( t ) = − 4.9 t 2 + 118 t + 115 .

Where a = -4.9, b = 118, c = 115

Let's assume that the trajectory of the rocket is a perfect parabola.

The time t the rocket will reach its maximum height will be at the symmetry of the parabola.

t = -b/2a

Substitute b and a into the formula

t = -118/-2(4.9)

t = 118/9.8

t = 12.041 seconds

Since NASA launches the rocket at t = 0 seconds, the time it will splash down into the ocean will be 2t.

2t = 2 × 12.041 = 24.08 seconds

Therefore, the rocket splashes down after 24.08 seconds.

B.) At maximum height, time t = 12.041s

Substitute t for 12.041 in the function

h ( t ) = − 4.9 t 2 + 118 t + 115

h(t) = -4.9(12.041)^2 + 118(12.041) + 115

h(t) = -4.9(144.98) + 118(12.041) + 115

h(t) = -710.402 + 1420.82 + 115

h(t) = 825.42 metres

Therefore, the rocket get to the peak at 825.42 metres

6 0

3 years ago

Use identities to find the values of the sine and cosine functions for the following angle measure.

puteri [66]

Using the cosine double angle formula,

$\cos 2\theta=2\cos^2 \theta-1=\frac{12}{13}\\\\2\cos^{2} \theta=\frac{25}{13}\\\\\cos^{2} \theta=\frac{25}{26}\\\\\boxed{\cos \theta=\frac{5}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

Using the Pythagorean identity,

$\sin^2 \theta+\cos^2 \theta=1\\\\\sin^2 \theta+\frac{25}{26}=1\\\\sin^2 \theta=\frac{1}{26}\\\\\boxed{\sin \theta=\frac{1}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

6 0

1 year ago

According to the latest survey the mean world lifespan is 68.8 years the university of Oregon wants to see if the average life s

pishuonlain [190]

Answer:

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

µ = 68.8

For the alternative hypothesis,

µ ≠ 68.8

This is a two tailed test.

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 26,

Degrees of freedom, df = n - 1 = 26 - 1 = 25

t = (x - µ)/(s/√n)

Where

x = sample mean = 72.8

µ = population mean = 68.8

s = samples standard deviation = 2.5

t = (72.8 - 68.8)/(2.5/√26) = 8.16

We would determine the p value using the t test calculator. It becomes

p < 0.00001

Assuming a significance level of 0.05, then

Since alpha, 0.05 > than the p value, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the average life span of the residence of Oregon differs from the world average.

4 0

3 years ago

Divide. 660 ÷ 3 The quotient is and the remainder is

posledela

$\bf{\blue{\underline{\underline{\pink{SOLUTION:-}}}}}$

=> 660 ÷ 3

=> 220

☃️ Quotient :- 220

☃️ Reminder :- 0

8 0

3 years ago

PLEASE HELP!

Aleks04 [339]

Answer:

Option c:

$0.25 \pm 1.645\sqrt{\frac{0.25*(1-0.25)}{90}}$

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

You have access to first year enrolment records and you decide to randomly sample 119 of those records. You find that 89 of those sampled went on to complete their degree. This means that $n = 119, \pi = \frac{89}{119} = 0.7478$ .