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Scorpion4ik [409]
2 years ago
7

A municipality has budgeted R80 000 for putting up new street name boards The street name boards cost R134 each. How many new st

reet name boards can be put up, and how much money will be left in the budget?​
Mathematics
1 answer:
ivanzaharov [21]2 years ago
6 0

The Municipality we be able to set up 597 street name boards with R2 left in the budget.

Data;

  • Amount Budgeted = R80,000
  • Cost of each board = R134

<h3>Number of Street Boards in the Budget</h3>

The number of streets boards that can be produced in the budget is calculated by dividing the total amount budgeted by the cost of each street board. This is done mathematically as

x = \frac{80000}{134}\\x = 597.01

We would have a total of 597 street names on the budget with some amount left.

We can calculated this by multiplying 597 by 134 and then subtracting the value from R80,000

597 * 134 = 79998 \\R80,000 - R79998 = R2

The Municipality we be able to set up 597 street name boards with R2 left in the budget.

Learn more on division of numbers here;

brainly.com/question/20301788

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The navy reports that the distribution of waist sizes among male sailors is approximately normal, with a mean of 32.6 inches and
grandymaker [24]

Answer:

a) 87.49%

b) 2.72%

Step-by-step explanation:

Mean of the waist sizes = u = 32.6 inches

Standard Deviation of the waist sizes = \sigma = 1.3 inches

It is given that the Distribution is approximately Normal, so we can use the z-distribution to answer the given questions.

Part A) A male sailor whose waist is 34.1 inches is at what percentile

In order to find the percentile score of 34.1 we need to convert it into equivalent z-scores, and then find what percent of the value lie below that point.

So, here x = 34.1 inches

The formula for z score is:

z=\frac{x-u}{\sigma}

Using the values in the above formula, we get:

z=\frac{34.1-32.6}{1.3}=1.15

Thus, 34.1 is equivalent to z score of 1.15

So,

P( X ≤ 34.1 ) =  P( z ≤ 1.15 )

From the z-table we can find the probability of a z score being less than 1.15 to be: 0.8749

Thus, the 87.49 % of the values in a Normal Distribution are below the z score of 1.15. For our given scenario, we an write: 87.49% of the values lie below 34.1 inches.

Hence, the percentile rank of 34.1 inches is 87.49%

Part B)

The regular measure of waist sizes is from 30 to 36 inches. Any measure outside this range will need a customized order.  We need to find that what percent of the male sailors will need a customized pant. This question can also be answered by using the z-distribution.

In a normal distribution, the overall percentage of the event is 100%. So if we find what percentage of values lie between 30 and 36, we can subtract that from 100% to obtain the percentage of values that are outside this range and hence will need customized pants.

First step is again to convert the values to z-scores.

30 converted to z scores will be:

z=\frac{30-32.6}{1.3}=-2

36 converted to z score will be:

z=\frac{36-32.6}{1.3}=2.62

So,

P ( 30 ≤ X ≤ 36 ) = P ( -2 ≤ z ≤ 2.62 )

From the z table, we can find P ( -2 ≤ z ≤ 2.62 )

P ( -2 ≤ z ≤ 2.62 ) = P(z ≤ 2.62) - P(z ≤ -2)

P ( -2 ≤ z ≤ 2.62 ) = 0.9956 - 0.0228

P ( -2 ≤ z ≤ 2.62 ) = 0.9728

Thus, 97.28% of the values lie between the waist sizes of 30 and 36 inches. The percentage of the values outside this range will be:

100 - 97.28 = 2.72%

Thus, 2.72% of the male sailors will need custom uniform pants.

The given scenario is represented in the image below. The black portion under the curve represents the percentage of male sailors that will require custom uniform pants.

3 0
4 years ago
A 30-foot tree casts a shadow that is 6 feet long. A person standing next to the tree casts a shadow that is 1.1 ft long. How ta
melomori [17]
The person is 5.5ft long
Reasoning: 30 divided by 6 equals 5
5 times 1.1 equals 5.5
7 0
2 years ago
Please answer quickly!!!
dezoksy [38]

Answer:

A.

Step-by-step explanation:

5 0
3 years ago
(b) 5.2m + 5 = 2.9m – 3 – 4.1m
ohaa [14]

Answer:

m =-1.25

Step-by-step explanation:

5.2m + 5 = 2.9m – 3 – 4.1m

Combine like terms

5.2m + 5 = -1.2m – 3

Add 1.2m to each side

5.2m +1.2m +5 = -1.2m +1.2m -3

6.4m +5 = -3

Subtract 5 from each side

6.4m +5-5 = -3-5

6.4m = -8

Divide each side

6.4m/6.4 = -8/6.4

m =-1.25

7 0
3 years ago
Orange ribbon 2 meters long. The blue ribbon was 4/5 as long ad the orange. Linda cuts off a piece of blue ribbon. The length of
masha68 [24]

The initial length of the blue ribbon = 1.6 meters

Step-by-step explanation:

Here, according to the question:

The initial length of orange ribbon = 2 meters

Let us assume the initial length of blue ribbon = S meters

Now, the length of blue ribbon = (4/5) times the length of orange ribbon

\implies S =( \frac{4}{5}) x Length of Orange ribbon

\implies S = (\frac{4}{5}) \times 2 = 1.6

or S  = 1.6 meters

Hence, the initial length of the blue ribbon is 1.6 meters.

8 0
3 years ago
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