6x3=18...............................................................................................

The front part of a tent had the dimensions shown in a diagram.What is the area of this part of the tent

attashe74 [19]

The area of this of the tent is 20

3 0

3 years ago

Read 2 more answers

Find the area of the polygon.

zalisa [80]

Answer:

$A=128\ cm^2$

Step-by-step explanation:

we know that

The area of the polygon is equal to the area of a square plus the area of four congruent triangles

so

$A=b^2+4[\frac{1}{2}(b)(h)]$

where

$b=8\ cm\\h=4\ cm$

substitute

$A=8^2+4[\frac{1}{2}(8)(4)]$

$A=64+64=128\ cm^2$

4 0

3 years ago

aalyn [17]

Answer:

Step-by-step explanation:

Using normal distribution,

z = (x - μ)/σ

μ= mean = 44 and

σ = standard deviation= 5.0

a) The probability that yield strength is at most 40=

P( x lesser than or equal to 40)

z = (40-44)/5= -0.8

Looking at the normal distribution table,

P( x lesser than or equal to 40) =0.2119

b) P(x greater than 62) = 1 - P(x lesser than or equal to 62)

z = (62-44)/5= 3.6

Looking at the normal distribution table,

P(x greater than 62) = 1 -0.99984

= 0.00016

c)P( 42 lesser than or equal to x lesser than or equal to 62)

= P(x lesser than or equal to 62) - P( x lesser than or equal to 40)

= 0.99987-0.2119= 0.78797

d) What yield strength value separates the strongest 75% from the others.

To get x for strongest 75, we get the z value corresponding to 0.75 from the table

z = 0.675= (x-44)/5

x = 3.375+44 = 47.375

The rest is 25% = 0.25

we get the z value corresponding to 0.25 from the table)

z = -0.67 = (x-44)/5

-3.35= x -44

x = -44+3.35= 40.65

yield strength value that separates the strongest 75% from the others

=47.375-40.65= 6.725

3 0

3 years ago

Assume that

JulsSmile [24]

Answer:

D - It is impossible to make a judgment with the given information.

Step-by-step explanation:

The fact that 1200 births were randomly selected and only 599 of such picks are girls does not give enough information on whether the birth is significantly high, low or neither. We must have other information to test for significance of the births proportion.

All we know is that;

Proportion of girls birth (p) = 599/1200 = 0.499. And by default, the proportion of male births (q) will be 1-p = 1-0.499 = 0.501.

If we examine the proportion closely, there seems to be no significant difference in the birth proportion.

Having said this, we cannot really imply that, the number of girls is significantly high. Or the number of girls is neither significantly low nor significantly high. Or the number of girls is significantly low.

The best subjective submission will be that, <em>there is no significant difference between girls birth and males birth.</em> The question of high or low (an alternative hypothesis) requires some further statistical test and this question does not provide further details.

8 0

3 years ago

The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds

Bumek [7]

Answer: 11.5%

Explanation:

Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds.

Now, we get the z-score for 720 seconds by the following formula:

$\text{z-score} = \frac{x - \mu}{\sigma}$

where

$t = \text{time for the auditor to finish his work } = 720 \text{ seconds}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}$

So, the z-score of 720 seconds is given by:

$\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \boxed{\text{z-score} = 1.2}$

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

$P(t \ \textgreater \ 720) 
\\ = P(z \ \textgreater \ 1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\ \boxed{P(t \ \textgreater \ 720) = 0.115}$

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice.

8 0

3 years ago