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Hatshy [7]
2 years ago
10

The body of a 154-pound person contains approximately 2×10^2 milligrams of gold and 6×10^3 milligrams of aluminum. Based on this

information, the number of milligrams of aluminum in the body is how many times greater than the number of milligrams of gold in the body?
can you show your work too?
re asked it cs some bozo stole my points but pls hellp asap
Mathematics
1 answer:
hram777 [196]2 years ago
3 0

Answer:

2x10^2 and 6x 10^

Step-by-step explanation:

Hope I helped UwU

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Write the Prime factors that 24 and 36 have in common By showing them multiplied together
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The prime factors that the numbers 24 and 36 have in common are found out to be; 2 × 2 × 3

<h3>How to get prime factors? </h3>

A prime factor of a number is simply factor of that number that is a prime number. This means that any of the prime numbers that can be multiplied to give the original number.

Now, we have;

Prime factors of 36; 2 × 2 × 3 × 3

Prime factors of 24; 2 × 2 × 2 × 3

The prime factors that they have in common will be; 2 × 2 × 3

Read more about Prime factors at; brainly.com/question/1081523

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Consuela earns a salary of 40000$ per year plus commission of 1000$ for each car she sells. Write and solve an equation that sho
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Which of the following is an equation in the form y=ax^2+bx+c of the parabola in the graph
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he blood platelet counts of a group of women have a​ bell-shaped distribution with a mean of 247.3 and a standard deviation of 6
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Answer:

The approximate percentage of women with platelet counts within 3 standard deviations of the​ mean is 99.7%.

Step-by-step explanation:

We are given that the blood platelet counts of a group of women have a​ bell-shaped distribution with a mean of 247.3 and a standard deviation of 60.7.

Let X = <em>t</em><u><em>he blood platelet counts of a group of women</em></u>

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

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

where, \mu = population mean = 247.3

           \sigma = standard deviation = 60.7

Now, according to the empirical rule;

  • 68% of the data values lie within one standard deviation of the mean.
  • 95% of the data values lie within two standard deviations of the mean.
  • 99.7% of the data values lie within three standard deviations of the mean.

Since it is stated that we have to calculate the approximate percentage of women with platelet counts within 3 standard deviations of the​ mean, or between 65.2 and 429.4, i.e;

         z-score for 65.2 =  \frac{X-\mu}{\sigma}

                                     =  \frac{65.2-247.3}{60.7}  = -3

         z-score for 429.4 =  \frac{X-\mu}{\sigma}

                                       =  \frac{429.4-247.3}{60.7}  = 3

So, it means that the approximate percentage of women with platelet counts within 3 standard deviations of the​ mean is 99.7%.

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