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3 years ago

If my assignment is worth 14% of my quiz weight which is 27% of my total weight what is the weight of that assignment

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

3.785%

Step-by-step explanation:

Assignment is 14% of the quiz

=> Assignment = 14% of the quiz

Quiz is 27% of the total weightage

=> 14% of 27%

Find 14% of 27%

14% of 27%

= 0.14 x 27 = 3.785%

The assignment is 3.785% of the total weightage.

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The value of a car decreases $500 for every 1,000 miles it is driven. The current value of the car is $23,000, and the car is dr

Hoochie [10]

Answer: C

Step-by-step explanation:

If we know the value of the car decreases $500 for every 1,000 miles, and that the car is driven about 10,000 miles every year, that means that you need to take the total value of the car (23,000) and subtract it from the amount of money it is losing per year. Again, the car is driven about 10,000 miles per year, so that means that the car will most likely continue to be driven 10,000 miles per year. If you do the math, for one year, the value of the car will drop $5,000 ($500 x 10, because it is $500 per every 1,000 miles) So, for each year, you can just multiply the number of years by $5,000 to find out how much the vehicle has depreciated over time.

Hope this helped you and made sense! Feel free to ask me any questions you have!

4 0

3 years ago

The width of a rectangle is 3 cm less than the length. What is the length if the width is 11 cm

Rudik [331]

Answer:

Step-by-step explanation:

i added it

7 0

2 years ago

A builder wants to build the bridge whose cross section is shown in the diagram. Two companies offer simple bids on building the

leva [86]

The length of the bridge is the distance from the beginning to the end.

The distance b between each beam is 9ft.

Let:

$I \to$ I-Beam

$d_I \to$ distance between I beam and the bridge

$b \to$ distance between each I beam

Given that:

$I = \frac 34 ft\\$

$d_I = 3 ft$

$Length = 55ft\ 6in$ --- length of the bridge

From the diagram (see attachment), there are: 6 I-beams.

So, the length of the 6 I-beams is:

$L_1 = 6 \times I$

$L_1 = 6 \times \frac 34$

$L_1 = \frac {18}4$

$L_1 = 4.5ft$

There are 2 I-beams beside the bridge

So, the distance between the 2 I-beams and the bridge is:

$d_1 =2 \times d_I$

$d_1 =2 \times 3ft$

$d_1 =6ft$

There are 5 spaces between the I-beams

So, the length of the total spaces is:

$L_2 = 5 \times b$

$L_2 = 5b$

The total length is:

$Length = L_1 + d_1 + L_2$

So, we have:

$4.5ft + 6ft + 5b = 55ft\ 6in$

Collect like terms

$5b = 55ft\ 6in - 4.5ft - 6ft$

$5b = 44.5ft\ 6in$

Convert inches to feet

$5b = 44.5ft\ + \frac{6}{12}ft$

$5b = 44.5ft\ + 0.5ft$

$5b = 45ft$

Divide both sides by 5

$b = 9ft$

Hence, the distance (b) between each beam is 9ft.

Read more about lengths at:

brainly.com/question/22059747

3 0

3 years ago

Maria has 50 Star stickers, 33 Moon stickers, and 11 Earth stickers.

alexandr1967 [171]

Answer:

94 stickers

Step-by-step explanation:

50 + 33 + 11 = 94

8 0

3 years ago

Read 2 more answers

Choose an American adult at random. The probability that you choose a woman is 0.52 . 0.52. The probability that the person you

Nikitich [7]

Answer:

The probability that the person you choose is either a woman or has never been married (or both) is therefore about 0.66

Step-by-step explanation:

Given Data:

Probability of choosing a women = P(W) = 0.52

Probability that chosen person has never married = P(M) = 0.26

Probability of choosing a women that has never married = P(W and M) = 0.11

We need to find the probability that the person you choose is either a woman or has never been married (or both). The addition rule of probabilities of two events is given as:

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) is the probability of occurrence of either A or B or both. P(A and B) is the probability of occurrence of A and B at the same time.

Re-writing the addition formula for given case, we get:

P(W or M) = P(W) + P(M) - P(W and M)

Substituting the given values results in:

P(W or M) = 0.52 + 0.26 - 0.11

P(W or M) = 0.67

Therefore, the probability that the person you choose is either a woman or has never been married (or both) is 0.67.

Note: There is either typo in given options or the questions. The question I found on Google has 0.25 as the probability that the person you choose has never married ( i.e. P(M) = 0.25 ).

So, in this case the correct answer would be option (b) 0.66

4 0

3 years ago