All categories

3 years ago

Please help me with my homework please

Mathematics

2 answers:

Wewaii [24]3 years ago

8 0

Solution:

1)Distance covered by cyclist=3.75 miles.

Time =0.3 hours.

Speed= Distance ÷Time

Speed =3.75÷0.3=12.5 miles per hour.

2) Speed = 12.5 miles per hour, Distance = 4.5 miles .

Time = Distance ÷Speed.

Time = 4.5÷12.5=0.36 hours.

Time taken to cover 4.5 miles is 0.36 hours.

mylen [45]3 years ago

7 0

Answer:

a) Speed of the cyclist is (2.75/0.4) = 6.875 miles per hour.

b) It will take (5.5/6.875) = 0.8 hours to travel 5.5 miles at this rate.

Remember: Speed=Distance/Time

You might be interested in

Given the functions f (x) = x^2 + 4 and g (x) = 3x+ 8 what is a solution for the equation ? x^2 + 4 = 3x + 8 A. 0 B. 1 C. 2 D. 4

k0ka [10]

Answer:

Step-by-step explanation:

Let's solve your equation step-by-step.

x2+4=3x+8

Step 1: Subtract 3x+8 from both sides.

x2+4−(3x+8)=3x+8−(3x+8)

x2−3x−4=0

Step 2: Factor left side of equation.

(x+1)(x−4)=0

Step 3: Set factors equal to 0.

x+1=0 or x−4=0

x=−1 or x=4

Answer:

x=−1 or x=4

6 0

3 years ago

What is 85/20 simplified

inessss [21]

Answer: 4 1/4

Step-by-step explanation: 85 divided by 20 is 4.25

6 0

3 years ago

Read 2 more answers

At sunrise, the temperature was –4.5°C. During the day, the temperature rose 9.7°C and then fell 4.8°C.

ollegr [7]

-4.5 rose 9.7 then fell 4.8

-4.5 + 9.7 = 5.2

5.2 - 4.8 = 0.4

8 0

3 years ago

The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.

AfilCa [17]

Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

Step-by-step explanation:

Let the random variable X denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

X = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

(a)

The formula to compute the sample variance for the daily price change is:

$s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}$

The sample mean is computed using the formula:

$\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

$\bar X$ =AVERAGE(A2:A13)

$s^{2}$ =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

$s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

$s$ =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

(c)

The (1 - α)% confidence interval for population variance is:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

Compute the critical value of Chi-square for α = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816$

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

$=[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]$

$=[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]$

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

7 0

4 years ago

Please help :( It’s due

jeyben [28]

Answer: 28

Step-by-step explanation: 28

5 0

1 year ago