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andreyandreev [35.5K]
3 years ago
13

A man six foot two inches tall casts a nine foot three inch shadow. How tall is a child that casts a five foot four inch shadow

at the same time of the day? (to the nearest whole inch)
Mathematics
1 answer:
saw5 [17]3 years ago
5 0

It will be a an 8 foot shadow

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Of the total number of spectators at a circus show, 1/4 are men. 2/5 of the remaining number of spectators are women. There are
ch4aika [34]

Answer: There are 198 children at the circus show.

Step-by-step explanation:

Let be "x" the total number of spectators at the circus show and "c" the number of children at the circus show.

Knowing that \frac{1}{4} of the spectators are men, we can find the remaining:

x-\frac{1}{4}x=\frac{3}{4}x

Since \frac{2}{5} of the remaining number of spectaros are women and there are a total of 132 women, we can write the following equation:

(\frac{2}{5})(\frac{3}{4}x)=132

Solving for "x", we get:

x=(132)(\frac{20}{6})\\\\x=440

Therefore, we get that "c" is:

 c=440-\frac{1}{4}(440)-132\\\\c=198

8 0
3 years ago
Find the measure of each angle please
Salsk061 [2.6K]
B is 80
C and D are 50
8 0
3 years ago
11/28+4/7 = simplest form
MatroZZZ [7]
The answer to this question is 27/28
4 0
3 years ago
The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a
AlekseyPX

Answer:

The time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

Step-by-step explanation:

We are given that the time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a standard deviation of 40 seconds.

The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition.

<u><em>Let X = time for this event for boys in secondary school</em></u>

SO, X ~ Normal(\mu=460,\sigma^{2} =40^{2})

The z-score probability distribution for normal distribution is given by;

                               Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean time = 460 seconds

            \sigma = standard deviation = 40 seconds

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

<u>Now, it is given that the fitness association wants to recognize the fastest 10% of the boys with certificates of recognition, which means;</u>

       P(X > x) = 0.10   {where x is the required time which boy need to beat}

       P( \frac{X-\mu}{\sigma} > \frac{x-460}{40} ) = 0.10

        P(Z > \frac{x-460}{40} ) = 0.10

<em>So, the critical value of x in the z table which represents the top 10% of the area is given as 1.2816, that is;</em>

<em>                   </em>      \frac{x-460}{40} =1.2816

                         {x-460}{} =1.2816\times 40

<em>                           </em>x = 460 + 51.264 = <u>511.264 seconds</u>

Hence, the time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

8 0
3 years ago
How would you "remove the discontinuity" of f ? in other words, how would you define f(5) in order to make f continuous at 5?f(x
Nataliya [291]
Hope this helps! I'm sorry for worse handwriting.

7 0
3 years ago
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