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3 years ago

Can anyone help me out plz

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

4 0

Negative rate of change just means that the graph is falling. On this one, that's everywhere after x=0.5. The only choice that's completely in that section is the last one.

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A person whose eye level is 20 feet above the ground is looking at a building. The angle of elevation to the top of the building

Inessa [10]

Distance of the person from the building (x)

tan 24 = 20/x => x = 20/tan 24 = 44.92 ft

The height of the building above the height of the person and the distance of the person from the building forms two legs of right angled triangle.

Therefore, if h is the height of the building, then;
tan 31 = (h-20)/44.92
h-20 = 44.92 tan 31
h = (44.92 tan 31) + 20 = 26.99 + 20 = 46.99 ft

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3 years ago

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Answer:

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3 years ago

Which box plot represents a set of data that has the least mean absolute deviation?

steposvetlana [31]

Answer:

try the second one because it is the smallest and close togehet so maybe that means that the mean is closer to the number and it nots some outlier.

Step-by-step explanation:

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3 years ago

Read 2 more answers

An employee earns $28.50 after 3 hours and $237.50 after 25 hours. What is the slope? What does the slope represent?

konstantin123 [22]

Answer:

You ad $28.50 and $237.50 and ad 25 and 3 so 266.00 divide into 28.

5 0

3 years ago

The average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 in

Murrr4er [49]

Answer:

Probability that average height would be shorter than 63 inches = 0.30854 .

Step-by-step explanation:

We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

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

where, $\mu$ = population mean = 64 inches

$\sigma$ = standard deviation = 4 inches

n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{63-64}{\frac{4}{\sqrt{4} } }$ ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

= 1 - 0.69146 = 0.30854

Hence, it is 30.85% likely that average height would be shorter than 63 inches.

7 0

3 years ago