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4 years ago

Someone help please

Mathematics

2 answers:

Masteriza [31]4 years ago

7 0

Answer: C) 42.3 cm^2

formula for the area if a trapzoid:
A= .5*(b1+b2)*h

A= .5(5.6+8.5)*6
A= 42.3

Vadim26 [7]4 years ago

7 0

42.3cmjdjsusjdhjdjdjdjdjhddjekiddjdjdjdjdjdjdididdiidjddijdjdjdiidi

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Write a number that is greater than 8.604 but less than 8.643

Aliun [14]

8.605
This number is bigger than 6.604 and it is smaller than 8.643
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6 0

3 years ago

Read 2 more answers

Help, will give brainlist!!

Licemer1 [7]

Answer:

Fred.

Started hang gliding at a height of 700 ft and descends 15 feet every seconds

Gene

Started hang gliding at a height of 575 ft and descends 10 feet every seconds

Step-by-step explanation:

The function that models Fred's hang gliding is $f(x)=-15x+700$

The initial value is 700 feet. This Fred was 700 feet above see level before he starts descending.

The rate of descent is -15 ft/s. This means Fred descends 15 feet in one second.

From the table the initial height is 575 ft. This means Gene was 575 feet above sea-level at the beginning of the hang gliding.

The rate of descent is $\frac{565-575}{1-0} =-10$ ft/s.

This means that in every seconds, Gene descends 10 feet.

4 0

4 years ago

An air transport association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maxim

Flura [38]

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the population mean, when the population standard deviation is not provided is:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

The sample selected is of size, n = 50.

The critical value of t for 95% confidence level and (n - 1) = 49 degrees of freedom is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, 49}\approx t_{0.025, 60}=2.000$

*Use a t-table.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552$

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

$=6.76\pm 2.000\times \frac{2.552}{\sqrt{50}}\\\\=6.76\pm 0.722\\\\=(6.038, 7.482)\\\\\approx (6.0, 7.5)$

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

7 0

3 years ago

Can someone help me with this math homework please!

statuscvo [17]

Answer:

1 = Input G

2 = Input F

3 = Input H

Step-by-step explanation:

5 0

3 years ago

A softball pitcher has a 0.675 probability of throwing a strike for each pitch and a 0.325 probability of throwing a ball. If th

Dafna11 [192]

Answer:

The probability that exactly 19 of them are strikes is 0.1504

Step-by-step explanation:

The binomial probability parameters given are;

The probability that the pitcher throwing a strike, p = 0.675

The probability that the pitcher throwing a ball. q = 0.325

The binomial probability is given as follows;

$p(x) = _{n}C_{x}\cdot p^{x}\cdot q^{1-x}$

Where:

x = Required probability

Therefore, the probability that the pitcher throws 19 strikes out of 29 pitches is found as follows;

The probability that exactly 19 of them are strikes is given as follows;

$\binom{29}{19}(0.675)^{19}0.325^{10} = \frac{29!}{19!\times 10!}\times (0.675)^{19}\times 0.325^{10} = 0.1504$

Hence the probability that exactly 19 of them are strikes = 0.1504

8 0

3 years ago