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Digiron [165]
2 years ago
14

Helpppppppppppppppppppppppppp

Mathematics
1 answer:
avanturin [10]2 years ago
5 0

Answer:

1/125, 1/625, 1/3125

Step-by-step explanation:

a8=25*(1/5)^7=25/78125=1/3125

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Please help and thank you.
Ulleksa [173]

Answer:

zeros: x = 5, or (5, 0)

domain: x ≥ -4

Step-by-step explanation:

The zeros are the values of x where the graph crosses the x-axis. The graph crosses at x=5, so that is the zero.

___

The domain is the set of x-values for which the function is defined. There is no graph for x < -4, so the graph is only defined for x ≥ -4. The domain is x ≥ -4.

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The graph has the appearance of the graph of ...

y=\sqrt{x+4}-3

8 0
3 years ago
Slope-intercept from by steps
nordsb [41]
Y=m+b use this equation
3 0
3 years ago
Assessment started: undefined.
SashulF [63]

Answer:

(x+4)

Step-by-step explanation:

6 0
3 years ago
The mean age of 5 people in a room is 36 years. A person enters the room. The mean age is now 39. What is the age of the person
____ [38]
Your answer is 36.5. Hope this helps!
7 0
3 years ago
The company states that 90% of the light bulbs have a lifetime of at least k hours. Find the value of k. Give your answer correc
docker41 [41]

Answer:

a) Mean = 5800 hours

b) P(5000 ≤ L ≤ 6000) = 0.420

c) k = 4700 hours

Step-by-step explanation:

The full correct question is attached to this solution.

From the graph of the distribution of lifetime of bulbs presented, it is evident that this is a normal distribution.

(a) Write down the mean lifetime of the light bulbs.

From the normal distribution graph, the mean is at the very centre of the graph

Mean = μ = 5800 hours

(b) The standard deviation of the lifetime of the light bulbs is 850 hours.

Find the probability that 5000 ≤ L ≤ 6000, for a randomly chosen light bulb.

Mean = μ = 5800 hours

Standard deviation = σ = 850 hours.

The required probability = P(5000 ≤ L ≤ 6000)

To find this, we first normalize/standardize 5000 and 6000.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 5000

z = (L- μ)/σ = (5000 - 5800)/850 = -0.941

For 6000

z = (L- μ)/σ = (6000 - 5800)/850 = 0.235

The required probability

P(5000 ≤ L ≤ 6000) = P(-0.941 ≤ z ≤ 0.235)

We'll use data from the normal probability table for these probabilities

P(5000 ≤ L ≤ 6000) = P(-0.941 ≤ z ≤ 0.235)

= P(z ≤ 0.235) - P(z ≤ -0.941)

= 0.593 - 0.173 = 0.420

c) The company states that 90% of the light bulbs have a lifetime of at least k hours. Find the value of k. Give your answer correct to the nearest hundred.

P(L ≥ k) = 90% = 0.9

assuming that the z-score of k is z'

P(L ≥ k) = P(z ≥ z') = 0.90

P(z ≥ z') = 1 - P(z < z') = 0.9

P(z < z') = 0.1

Using the normal distribution table

z' = -1.282

z' = (k - μ)/σ

-1.282 = (k - 5800)/850

k = (-1.282 × 850) + 5800 = 4710.3 = 4700 hours to the nearest hundred.

Hope this Helps!!!

5 0
3 years ago
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