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3 years ago

Directions: Match the product and quotients estimates below with the correct expression on the left

Mathematics

1 answer:

prohojiy [21]3 years ago

8 0

Answer:

Answer is in attached image.

Step-by-step explanation:

Given the expressions, for which we have to find the estimates as per the expressions on the left.

The given expressions are:

1) 35 $\times$ 23

2) 132 $\div$ 168

3) 17.3 $\times$ 18.4

4) 999 $\div$ 208

5) 998 $\times$ 211

Here, we need to find the rounded off numbers.

35 can be rounded to 40 and 23 to 20.

Therefore, equivalent to 35 $\times$ 23 is 40 $\times$ 20

132 can be rounded to 130 and 168 to 170.

Therefore, equivalent to 132 $\div$ 168 is 130 $\div$ 170.

17.3 can be rounded to 17.0 and 18.4 to 18.0.

Therefore, equivalent to 17.3 $\times$ 18.4 is 17.0 $\times$ 18.0.

999 can be rounded to 1000 and 208 to 210.

Therefore, equivalent to 999 $\div$ 208 is 1000 $\div$ 210.

998 can be rounded to 1000 and 211 to 210.

Therefore, equivalent to 998 $\times$ 211 is 1000 $\times$ 210.

The solution can be found in the attached image as well.

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3 years ago

Read 2 more answers

The mean of a population is 74 and the standard deviation is 16. The shape of the population is unknown. Determine the probabili

bulgar [2K]

Answer:

$P(X > 75)= 0.35402$

$P(72 < X < 75 ) = 0.2529$

$P( X < 74.7) = 0.74041$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 74$

The population standard deviation is $\sigma = 16$

Considering question a

The sample size is n = 36

Generally the standard error of mean is mathematically represented as

$\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_{x} = \frac{16}{\sqrt{36} }$

=> $\sigma_{x} = 2.67$

Generally the probability that a random sample of size 36 yielding a sample mean of 75 or more is mathematically represented as

$P(X > 75) = P( \frac{X - \mu }{ \sigma_{x}} > \frac{75 - 74}{ 2.67 } )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

$P(X > 75) = P(Z > 0.3745 )$

From the z table the area under the normal curve representing 0.3745 to the right is

$P(Z > 0.3745 ) = 0.35402$

=> $P(X > 75)= 0.35402$

Considering question b

The sample size is n = 104

Generally the standard error of mean is mathematically represented as

$\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_{x} = \frac{16}{\sqrt{104} }$

=> $\sigma_{x} = 1.5689$

Generally the probability that a random sample of size 104 yielding a sample mean between 72 and 75 is mathematically represented as

$P(72 < X < 75 ) = P(\frac{72 - 74 }{1.5689} < \frac{X - \mu }{\sigma_{x}} < \frac{75 - 74 }{1.5689} )$

=> $P(72 < X < 75 ) = P(-1.275 < Z < 0.375 )$

=> $P(72 < X < 75 ) = P(Z < 0.375 ) - P(Z < -1.275)$

From the z table the area under the normal curve representing -1.275 to to the left is

$P(Z < -1.275) =0.10115$

=> $P(72 < X < 75 ) = 0.35402 - 0.10115$

=> $P(72 < X < 75 ) = 0.2529$

Considering question c

The sample size is n = 217

Generally the standard error of mean is mathematically represented as

$\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_{x} = \frac{16}{\sqrt{217} }$

=> $\sigma_{x} = 1.086$

Generally the probability that a random sample of size 217 yielding a sample mean of less than 74.7 is mathematically represented as

$P( X < 74.7) = P(\frac{X - \mu }{\sigma_x} < \frac{ 74.7 - 74 }{ 1.086 })$

=> $P( X < 74.7) = P(Z < 0.6446 )$

From the z table the area under the normal curve representing 0.6446 to to the left is

$P(Z < 0.6446 ) = 0.74041$

=> $P( X < 74.7) = 0.74041$

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\\\\\\
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