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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

sladkih [1.3K]

Answer:

10.17 seconds

Step-by-step explanation:

substitute y=0 into the equation

0=-16x²+153x+98

then use the quadratic formula

a = -16

b= 153

c = 98

Substituting values into the Quadratic equation, you get (Round the values to the nearest hundreth):

x₁= 10.17

x₂= -0.60

5 0

2 years ago

An elementary ntimesn scaling matrix with k on the diagonal is the same as the ntimesn identity matrix with _______ exactly one

Dafna11 [192]

Answer:

exactly one, 0's, triangular matrix, product and 1.

Step-by-step explanation:

So, let us first fill in the gap in the question below. Note that the capitalized words are the words to be filled in the gap and the ones in brackets too.

"An elementary ntimesn scaling matrix with k on the diagonal is the same as the ntimesn identity matrix with EXACTLY ONE of the (0's) replaced with some number k. This means it is TRIANGULAR MATRIX, and so its determinant is the PRODUCT of its diagonal entries. Thus, the determinant of an elementary scaling matrix with k on the diagonal is (1).

Here, one of the zeros in the identity matrix will surely be replaced by one. That is to say, the determinants = 1 × 1 × 1 => 1. Thus, it is a a triangular matrix.

3 0

3 years ago

The bus will leave in 7 minutes. what time will it leave?

hjlf

Answer:

1:52

Step-by-step explanation:

The clock reads 1:45 so simply add 7 minutes to that

3 0

2 years ago

Read 2 more answers

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

3 years ago

DOG A dog is leashed to the corner of a house. How much running area does the dog have? Explain how you found your answer.

Eduardwww [97]

Answer:

The area will be $\frac{3\pi r^{2} }{4}$ where r is the length of the leash.

Step-by-step explanation:

The area of a circle is pi times the radius squared. One forth of that area is not available for the dog to run: its the corner of the house. so what's left is 3/4 of the area of the circle. (3/4)πr²

Since the length of the leash is not given in the question, no specific Area can be answered. Once that value is substituted for r, the actual area can be calculated.

7 0

3 years ago