Two similar cylinders have surface areas of 24 cm² and 54 cm². The volume of the smaller cylinder is 16ñ cm³.

1 a-2a-{3a-(4a-5)} i need help plz and on this one to 13(5-t)-(5-t)-12(5-t)

KengaRu [80]

Answer:

1. -5

2. 0

Step-by-step explanation:

1. Distribute the Negative Sign:

=a−2a+−1(3a−(4a−5))

=a+−2a+−1(3a)+−1(−4a)+(−1)(5)

=a+−2a+−3a+4a+−5

Combine Like Terms:

=a+−2a+−3a+4a+−5

=(a+−2a+−3a+4a)+(−5)

Answer:

=−5

2. Distribute:

=(13)(5)+(13)(−t)+−5+t+(−12)(5)+(−12)(−t)

=65+−13t+−5+t+−60+12t

Combine Like Terms:

=65+−13t+−5+t+−60+12t

=(−13t+t+12t)+(65+−5+−60)

Answer:

=0

8 0

3 years ago

What is the slope of the line shown below? A. -5 B. 5 C.-2 D. 2

expeople1 [14]

Answer:

D. 2

Step-by-step explanation:

rise/run = $\frac{y_{2} -y_1}{x_{2} -x_1} }$ = $\frac{7 - -3}{3- -2}$ = 10/5 = 2

8 0

3 years ago

What’s the answer ????

butalik [34]

Answer:

A-1

Step-by-step explanation:

It perfectly represents the scenario.

4 0

3 years ago

Read 2 more answers

Aye plz help. Im abt to die. Find the perimiter of the shape below.

Law Incorporation [45]

Answer:

40

Step-by-step explanation:

Add all the sides together which is 40

4 0

3 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

3 years ago