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harkovskaia [24]
3 years ago
12

Write a for loop that computes the following sum: 5 10 15 20 ... 485 490 495 500. The sum should be placed in a variable sum tha

t has already been declared and initialized to 0. In addition, there is another variable, num that has also been declared. You must not use any other variables.

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
8 0

The solution is in the attachment

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Plz help plzzzzzzzzzzzz
saveliy_v [14]

Answer:

After 1.5 second of throwing the ball will reach a maximum height of 44 ft.

Step-by-step explanation:

The height in feet of a ball after t seconds of throwing is given by the function  

h = - 16t² + 48t + 8 .......... (1)

Now, condition for maximum height is  

\frac{dh}{dt} = 0 = - 32t + 48 {Differentiating equation (1) with respect to t}

⇒ t = \frac{48}{32} = 1.5 seconds.

Now, from equation (1) we get

h(max) = - 16(1.5)² + 48(1.5) + 8 = 44 ft.

Therefore, after 1.5 seconds of throwing the ball will reach a maximum height of 44 ft. (Answer)

4 0
3 years ago
Please please helppppppppppppp
Leya [2.2K]

Answer:

C

Step-by-step explanation:

Both lines start and end on x = 3

The lower line has a positive slope and a y- intercept of - 2

The lower line has an open circle at it's endpoint indicating it cannot equal 3

x - 2 for x < 3

The upper line has a positive slope and a y- intercept of - 1, if we project the line. The line has a closed circle at it's start indicating it can equal 3, thus

x - 1 for x ≥ 3

Hence

x - 1 if x ≥ 3

x - 2 if x < 3 → C

3 0
3 years ago
6.Madeline made the wooden steps shown. What is the volume of the steps? The wooden steps have two parts, both of which are rect
Lelu [443]

Answer:

840 cubic inches

Step-by-step explanation:

Given:

The wooden steps have two parts.

Both of which are rectangular prisms.

The top part is 6 inches tall, 5 inches wide and 10 inches long.

The bottom part is 6 inches tall, 9 inches wide and 10 inches long.

Question asked:

What is the volume of the steps?

Solution:

<u>For top part</u>

Length = 10 inches

Width = 5 inches

Height = 6 inches

<u><em>As we know:</em></u>

<u><em /></u>Volume\ of\ rectangular\ prism =length\times width\times height

                                                 =10\times5\times6=300\ cubic\ inches

<u>For bottom part</u>

Length = 10 inches

Width = 9 inches

Height = 6 inches

Similarly

Volume\ of\ rectangular\ prism =length\times width\times height

                                                 =10\times9\times6=540\ cubic\ inches

Volume of two steps = 300 + 540 = 840 cubic inches

<em>Therefore, the volume of the steps are 840 cubic inches.</em>

<em></em>

8 0
3 years ago
Find the quotient of <br><br> -85^2-27n+76 / -17n-19
bagirrra123 [75]
<u><em>Answer:</em></u>
5n-4

<u><em>Explanation:</em></u>
<u>The given expression is:</u>
\frac{-85n^2-27n+76}{-17n-19}

<u>1- Take the negative as a common factor from both the numerator and the denominator. This will give us:</u>
\frac{-(85n^2+27n-76)}{-(17n+19)}

<u>2- Cancel out the negative sign (common factor) from the numerator and denominator. This will give us:</u>
\frac{(85n^2+27n-76)}{(17n+19)}

<u>3- Factor the numerator. This wil give:</u>
\frac{(5n-4)(17n+19)}{(17n+19)}

<u>4- Finally, cancel out the (17n+19) which common in both numerator and denominator. This will give us the final expression:</u>
5n-4

Hope this helps :)
6 0
3 years ago
A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade
Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

P(x\geq1)=1-P(x

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly  n= 5

Number of blades in an assembly dull is M  = 10.

The probability mass function is,

P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

P(x\geq 1)=1- P(x

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly  N = 48

Number of blades selected at random from the assembly  N = 5

Number of blades in an assembly dull is  M = 10

From the information,

The probability that assembly is replaced (P)  is 0.7069.

The probability that assembly is not replaced is (Q)  is,

q=1-p\\= 1-0.7069= 0.2931

The geometric probability mass function is,

P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)

The probability that assembly is replaced no replaced until the third day of evaluation is,

P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly   N = 48

Number of blades selected at random from the assembly  n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 2.

P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 6.

P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

P(x\geq 1)=1- P(x

 

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0
3 years ago
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