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leva [86]
3 years ago
13

A baseball is dropped from a stadium seat that is 73 feet above the ground. Its height s in feet after t seconds is given by s(t

)=73−16t^2. Estimate how long it takes for the baseball to strike the ground.
Mathematics
1 answer:
lisov135 [29]3 years ago
8 0
0 = 73 - 16t^2 
16t^2 = 73
4^2*t^2 = 73
(4t)^2 = 73
<span>sqrt(4t^2) = sqrt(73) </span>
4t = 8.5440
<span>t = 2.136</span>
ANSWER: 2.1 seconds
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kumpel [21]
Answer:
x = 10

Explanation:
Note the following rules before we begin:
2 ln(a) = ln (a²)
ln (a) + ln (b) = ln (ab)

Now, for the given:
ln(20) + ln(5) = 2 ln(x)
ln(20*5) = ln(x²)
ln(100) = ln(x²)

This means that:
100 = x²
Therefore:
either x = 10 .........> acceped
or x = -10 .........> rejected as no negatives are allowed within ln functions

Hope this helps :)
5 0
3 years ago
In 25 words or fewer, compare the mathematical and algebraic<br><br> expressions.
DedPeter [7]

Answer:

use own words

Step-by-step explanation:

An expression is just a mathematical phrase. In this tutorial, you'll learn about two popular types of expressions: numerical and algebraic expressions. A numerical expression contains numbers and operations. An algebraic expression is almost exactly the same except it also contains variables.

in mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number)

5 0
3 years ago
A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 98% conf
adelina 88 [10]

Answer:

n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337

So the answer for this case would be n=1337 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=1.1 represent the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =0.07 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.01;0;1)", and we got z_{\alpha/2}=2.326, replacing into formula (b) we got:

n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337

So the answer for this case would be n=1337 rounded up to the nearest integer

3 0
4 years ago
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galben [10]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
A random sample of 250 students at a university finds that these students take a mean of 15.3 credit hours per quarter with a st
SpyIntel [72]

Answer:  15.3\pm0.198

OR

(15.102, 15.498)

Step-by-step explanation:

The formula to find the confidence interval(\mu) is given by :-

\overline{x}\pm z_{\alpha/2, df}\dfrac{s}{\sqrt{n}}

, where n is the sample size

s = sample standard deviation.

\overline{x}= Sample mean

z_{\alpha/2} = Two tailed z-value for significance level of \alpha .

Given : Confidence level = 95% = 0.95

Significance level = \alpha=1-0.95=0.05

s= 1.6

\overline{x}=15.3

sample size : n= 250 , which is extremely large ( than n=30) .

So we assume sample standard deviation is the population standard deviation.

thus , \sigma=1.6

By standard normal  distribution table ,

Two tailed z-value for Significance level of 0.05 :

z_{\alpha/2}=z_{0.025}=1.96

Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-

15.3\pm (1.96)\dfrac{1.6}{\sqrt{250}}\\\\ =15.3\pm 0.19833\\\\=\approx15.3\pm0.198\\\\=(15.3-0.198,\ 15.3+0.198)=(15.102,\ 15.498)

Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)

5 0
3 years ago
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