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aniked [119]
3 years ago
7

At Roosevelt High School, there are six more students on the

Mathematics
1 answer:
motikmotik3 years ago
6 0

Answer:

24 students work on the newspaper and 18 are on the boys' tennis team

Step-by-step explanation

42-6=36

36/2=18

18+6=24

24+18=42

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There are 575 fireworks to be shot off in a firework display every minute 12 new fireworks are shot off display write a verbal m
lukranit [14]

Complete question :

There are 575 fireworks to be shot off in a firework display every minute 12 new fireworks are shot off display write a verbal model and algebraic expression to represent the number of fireworks left to be shot off after t minutes.

Answer:

575 - 12t

Step-by-step explanation:

Given the following :

Total number of fireworks = 575

Number of shots per minute = 12

To calculate the number of fireworks left to be shot off after t minutes, The total number of fireworks already shot after the same time interval t in minutes, is first obtained, this is equivalent to (12*t). The result is then subtracted from the total number of fireworks to be shof off.

In algebraic terms

[total number of fireworks on display - (number of shots per minute × t)]

575 - 12t

8 0
3 years ago
Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally
rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = <u><em>thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.</em></u>

So, X ~ Normal(\mu=5.0,\sigma^{2} =1.7^{2})

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

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

where, \mu = population mean thickness = 5.0 mm

           \sigma = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

    P(X < 3.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{3.0-5.0}{1.7} ) = P(Z < -1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

    P(X > 7.0 mm) = P( \frac{X-\mu}{\sigma} > \frac{7.0-5.0}{1.7} ) = P(Z > 1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X \leq 3.0 mm)

    P(X < 7.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{7.0-5.0}{1.7} ) = P(Z < 1.18) = 0.881

    P(X \leq 3.0 mm) = P( \frac{X-\mu}{\sigma} \leq \frac{3.0-5.0}{1.7} ) = P(Z \leq -1.18) = 1 - P(Z < 1.18)

                                                           = 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

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4 years ago
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Answer:

Step-by-step explanation:

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the area of a circle is A= πr²

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