The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.

5 0

3 years ago

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Simplify 3xy + x 2 - 4xy + 2xupper2y

densk [106]

The answer is X^2-xy+2x^2y sorry if I am wrong

6 0

3 years ago

The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).

natita [175]

Answer:

68% of jazz CDs play between 45 and 59 minutes.

Step-by-step explanation:

<u>The correct question is:</u> The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).

According to the 68-95-99.7 rule, what percentage of jazz CDs play between 45 and 59 minutes?

Let X = <u>playing time of jazz CDs</u>

SO, X ~ Normal( $\mu=52, \sigma^{2} =7$ )

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

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

Now, according to the 68-95-99.7 rule, it is stated that;

68% of the data values lie within one standard deviation points from the mean.

95% of the data values lie within two standard deviation points from the mean.

99.7% of the data values lie within three standard deviation points from the mean.

Here, we have to find the percentage of jazz CDs play between 45 and 59 minutes;

For 45 minutes, z-score is = $\frac{45-52}{7}$ = -1

For 59 minutes, z-score is = $\frac{59-52}{7}$ = 1

This means that our data values lie within 1 standard deviation points, so it is stated that 68% of jazz CDs play between 45 and 59 minutes.

6 0

3 years ago

Which equivalent four-term polynomial can be created

Korvikt [17]

Answer:

x2 + 12x + 4x + 48

Step-by-step explanation:

i just took the assignment

6 0

3 years ago