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2 years ago

Kali left school and traveled toward her friend’s house at an average speed of 40

1 answer:

ioda2 years ago

3 0

Matt needs to travel 3 hours.
If 40km=1 hour
400km=? But there are two distances
200/40*1=5 hours
Therefore Kali needs to travel 5 hours to be 200 km away from the house.
If 50km=1 hour
200km=?
200/50*1hr=4hrs Therefore Matt needs to travel 4hrs to be 200 km away from the house. But Kali traveled 1hr earlier than Matt. 4hrs-1hr=3hrs
Therefore Matt has to travel for 3 hours to be 400km away from Kali.

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Triangle A B C is shown. Angle A C B is a right angle. The length of hypotenuse A B is 12 centimeters, the length of C B is 9.8

Paha777 [63]

Answer:

tan−1(StartFraction 6.9 Over 9.8 EndFraction)

Step-by-step explanation:

tan−1(StartFraction 6.9 Over 9.8 EndFraction)

tan = opp/adj = 9.8/6.9

tan -1 = 1 / tan = 1 / (9.8 /6.9) = 6.9 /9.8

4 0

1 year ago

Read 2 more answers

A yard is equal in length to three feet. The function f(x) takes a measurement in yards (as input) and returns a measurement in

Grace [21]

If it takes yards and turns it into feet then it just multiplies it by 3

so if the input is 10.5 then you just multiply it by three

31.5 feet is the imput

3 0

1 year ago

Read 2 more answers

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago

Leah loves chicken wings and is comparing the deals at three different restaurants. Buffalo Bills has 888 wings for \$7$7dollar

ella [17]

Cost of chicken wings at Buffalo Bills = 8 wings for $7

Cost of 1 wing at Buffalo Bills = $\frac{7}{8} =0.875$

Cost of chicken wings at Buffalo Mild Wings = 12 wings for $10

Cost of 1 wing at Buffalo Mild Wings = $\frac{10}{12}= 0.833$

Cost of chicken wings at Wingers = 20 wings at $17

Cost of 1 wing at Wingers = $\frac{17}{20}= 0.850$

Hence, comparing all the three costs per wing, we can see that Buffalo Mild Wings is serving chicken wings at lowest price of $0.833 per wing.

8 0

1 year ago

Read 2 more answers

“there are several approximations used for pi, including 3.14 and 22/7. pi is approximately 3.14159265358979..." a. label pi and

BlackZzzverrR [31]

Hello,

a) | |PA|-π |>| |PB|-π | (see pic)

c) 3.14< x/113 <=π <22/7

==>3.14*113<x<22/7 * 113
==>354.82 <x < 355.14285
==>x=355

Answer 355/113=3,1415929203539823008849557522124

6 0

1 year ago