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liraira [26]
3 years ago
11

A seamstress uses 12 yards of fabric to make 3 costumes for students in the chorus. How many yards will she need to make costume

s for all 25 students in chorus?
Mathematics
2 answers:
Step2247 [10]3 years ago
7 0

Answer:

100 yards

Step-by-step explanation:

It takes 4 yards of fabric to make a costume for one student.  She needs to make costumes for every student in chorus, so you just multiply this number by 25 and get 100 yards.

AleksAgata [21]3 years ago
6 0

Answer:

100 yards.

Step-by-step explanation:

We have been given that a seamstress uses 12 yards of fabric to make 3 costumes for students in the chorus.

We will use proportions to solve the given problem.

\frac{\text{Yards of fabric}}{\text{Costumes}}=\frac{12}{3}

\frac{\text{Yards of fabric}}{25}=\frac{12}{3}

\frac{\text{Yards of fabric}}{25}*25=\frac{12}{3}*25

\text{Yards of fabric}=4*25

\text{Yards of fabric}=100

Therefore, 100 yards of fabric is needed to make costumes for 25 students.

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|14+x|-5=k solve for k​
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Step-by-step explanation:

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6 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.

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