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Vikentia [17]
2 years ago
7

A stadium asked for $760,000 for naming rights. During a bidding war for the stadium naming rights, one company bid 7% more than

the amount requested, and a second company bid 2% more than what that company bid. How much did the second company bid
Mathematics
1 answer:
alex41 [277]2 years ago
8 0

Answer:

\$\ 829464

Step-by-step explanation:

Bid of first company:

Bid is 7\% more

7\% of 760000=\frac{7}{100}\times760000=53200

Bid of first =760000+53200=813200

Bid of second company:

Bid of second is 2\% greater than that of first

2\% of first bid =\frac{2}{100}\times 813200=16264

Hence Bid of second company =813200+16264=829464

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A study was performed on green sea turtles inhabiting a certain area.​ Time-depth recorders were deployed on 6 of the 76 capture
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Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.  

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

                    P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean shell length = 51.3 cm

             s = sample standard deviation = 6.6 cm

             n = sample of turtles = 6

             \mu = true mean shell length

Now, the 99% confidence interval for \mu =  \bar X \pm t_(_\frac{\alpha}{2}_)  \times \frac{s}{\sqrt{n} }

Here, \alpha = 1% so  (\frac{\alpha}{2}) = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> \mu  =  51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }

                                                         = [51.3 - 10.864 , 51.3 + 10.864]

                                                         = [40.44 cm, 62.16 cm]

The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

3 0
2 years ago
In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, AE=x2−16 , and CE=6x .
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Answer:

<h3>AC=96 units.</h3>

Step-by-step explanation:

We are given a parallelogram ABCD with diagonals AC and BD intersect at point E.

AE=x^2-16. , and CE=6x .

<em>Note: The diagonals of a parallelogram intersects at mid-point.</em>

Therefore, AE = EC.

Plugging expressions for AE and EC, we get

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Subtracting 6x from both sides, we get

x^2-16-6x=6x-6x

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Factoriong quadratic by product sum rule.

We need to find the factors of -16 that add upto -6.

-16 has factors -8 and +2 that add upto -6.

Therefore, factor of x^2-6x-16=0 quadratic is (x-8)(x+2)=0

Setting each factor equal to 0 and solve for x.

x-8=0  => x=8

x+2=0  => x=-2.

We can't take x=-2 as it's a negative number.

Therefore, plugging x=8 in EC =6x, we get

EC = 6(8) = 48.

<h3>AC = AE + EC = 48+48 =96 units.</h3>
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