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jekas [21]
3 years ago
6

Please help attachment

Mathematics
2 answers:
larisa86 [58]3 years ago
4 0

Answer:

Step-by-step explanation:

a , 5\frac{9}{14}

b,7\frac{1}{2}

Artyom0805 [142]3 years ago
3 0

Answer:

A) 5 9/14

B) 7 1/2

Step-by-step explanation:

Question A asks what is 4 1/7 + 1 1/2

First turn the problem into improper fraction for it to be easier to calculate.

29/7 + 3/2

Now we have to find the denominator(bottom number)

We have 7 and 2 as our denominators. So we need to find out the lcm(least common factor)

For 7 and 2 the lcm is 14

So far we have:

29/7 + 3/2 = ?/14 + ?/14 = ?

Now we need to find the numerator(The top number). That fills in the ? over the two 14's.

To find the numerator we need to mutiply.

29/7 + 3/2 = ?/14 + ?/14 = ?

7 × ? = 14 the answer is 2 so then we multiply the top also with 2. So 29 × 2 = 58.

29/7 + 3/2 = 58/14 + ?/14 = ?

Now we have figured out one of the ?

We do the same thing for the other one

2 × ? = 14 answer is 7 so we also multiply the top with 7. 3 × 7 = 21.

29/7 + 3/2 = 58/14 + 21/14 = ?

Now we have to solve this:

58/14 + 21/14 = ?

The denominator is 14 because both of them are 14.

The top we need to add together so 58+21= 79

Now our answer is:

79/14

If we simplify it(by turning into improper fraction):

5 9/14

B) question is 4 1/2 ÷ 3/5

First we turn it into inproper fraction

9/2 ÷ 3/5

Since it is divsion we have to flip the second fraction which is 3/5 into 5/3, and turn the ÷ symbol into ×

9/2 × 5/3

Now we multiply across

9×5 = 45

2×3 = 6

answer: 45/6

Now if we simplify:

7 1/2

Hope that helped! that was long.. I would apreciate a brainlist if you could, thx!

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(a) the probability that an individual distance is greater than 214.80 cm is 0.1401.

(b) The probability that the mean for 15 randomly selected distances is greater than 204.00 cm is 0.2482.

(c) The normal distribution can be used because the original population has a normal distribution.

Step-by-step explanation:

We are given that the overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 cm.

(a) Let X = <em>the overhead reach distances of adult females</em>.

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

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

where, \mu = population mean reach distance = 205.5 cm  

           \sigma = standard deviation = 8.6 cm

So, X ~ Normal(\mu=205.5,\sigma^{2} =8.6^{2})

Now, the probability that an individual distance is greater than 214.80 cm is given by = P(X > 214.80 cm)

    P(X > 214.80 cm) = P( \frac{X-\mu}{\sigma} > \frac{214.80-205.5}{8.6} ) = P(Z > 1.08) = 1 - P(Z \leq 1.08)

                                                                    = 1 - 0.8599 = <u>0.1401</u>

The above probability is calculated by looking at the value of x = 1.08 in the z table which has an area of 0.8599.

(b) Let \bar X = <em>the sample mean selected distances</em>.

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

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

where, \mu = population mean reach distance = 205.5 cm  

           \sigma = standard deviation = 8.6 cm

           n = sample size = 15

Now, the probability that the mean for 15 randomly selected distances is greater than 204.00 cm is given by = P(\bar X > 204.00 cm)

    P(\bar X > 204 cm) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{204-205.5}{\frac{8.6}{\sqrt{15} } } ) = P(Z > -0.68) = 1 - P(Z \leq 0.68)

                                                                    = 1 - 0.7518 = <u>0.2482</u>

The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.7518.

(c) The normal distribution can be used in part​ (b), even though the sample size does not exceed​ 30 because the original population has a normal distribution and the sample of 15 randomly selected distances has been taken from the population itself.

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