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2 years ago

What fraction is represented by point A on the number line shown?

Mathematics

1 answer:

Archy [21]2 years ago

5 0

The fraction that is represented by point A of the number line shown is $1\frac{1}{3}$

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If g (x) = 2x + 2, find g (a + h) - g <br> a. .<br> a. 2h<br> b. 2h + 4<br> c. h + 2

katrin [286]

G(x) = 2x + 2

g(a + h) - g(a) = 2(a+h) + 2 - (2(a) + 2)

g(a + h) - g(a) = 2a + 2h + 2 - 2a - 2

g(a + h) - g(a) = 2h + 2 - 2

g(a + h) - g(a) = 2h

Your final answer is a. 2h.

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3 years ago

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Pump-a-Bunch Water Pump Company has a sump pump that will pump water with a flow of 2500 gallons per hour. Is the number of gall

Alexxx [7]

Answer:

Step-by-step explanation:

because

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3 years ago

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Philip’s rectangular deck is 5 yards wide. The ratio of the width to the length is 4 to 5. What is the approximate length of Phi

VladimirAG [237]

Answer:

12 feet

Step-by-step explanation:

just convert 4 yards into feet

8 0

3 years ago

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The volume of this cone is 3,183.96 cubic meters. What is the height of this cone?

Rom4ik [11]

Answer:

height = 18m

Step-by-step explanation:

cone volume = 3183.96 m³

1/3 πr²h = 3183.96

1/3×3.14×13×13×h = 3183.96

h = 3183.96×3/13×13×3.14

h = 9551.88/530.66

h = 18

5 0

2 years ago

Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.

lesya [120]

Answer:

0.1994 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851$

P(sample of 20 boxers is more than 167 pounds)

$P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)$

$= 1 - P(z \leq 0.8438)$

Calculation the value from standard normal z table, we have,

$P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%$

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0

3 years ago