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2 years ago

A city currently has 153 streetlights. As part of a urban renewal program, the city council has decided to

Mathematics

1 answer:

kiruha [24]2 years ago

5 0

Answer:

The explicit function is: $P_x = 153+3x$

And the number of lights after 33 weeks will be 252.

Step-by-step explanation:

Given that:

Total street lights = 153

Let x be the number of weeks

Then the number lights after x weeks will be 3x

This is a linear function where the y-intercept is 153 and slope is 3.

It can be written as: y = mx+b

The function is:

$P_x = 153+3x$

Putting the values for x will give us the number of total lights after that number of weeks.

To find, how many street lights were there at the end of 33rd week,

Putting x = 33

$P_{33} = 153+ 3(33)\\= 153+99 = 252$

Hence.

The explicit function is: $P_x = 153+3x$

And the number of lights after 33 weeks will be 252.

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Please help!

Aleks [24]

Answer:

Probability of picking an even and then even number = $\frac{1}{6}$

Step-by-step explanation:

Probability of picking a card with even number,

P(even 1) = $\frac{\text{favorable outcomes}}{\text{Total outcomes}}$ = $\frac{\text{even cards}}{\text{Total number of cards}}$

= $\frac{2}{4}$

= $\frac{1}{2}$

Followed by picking a card which is even without putting the first card back.

So an even card is to be picked out of remaining 3 cards.

P(even 2) = $\frac{\text{favorable outcomes}}{\text{Total outcomes}}$

= $\frac{1}{3}$

Now probability of both events = P(even 1) × P(even 2)

= $\frac{1}{2}\times \frac{1}{3}$

= $\frac{1}{6}$

Therefore, probability of both the events will be $\frac{1}{6}$ .

5 0

3 years ago

A certain company makes three grades (A, B, and C) of a particular electrical component. Historically, grade A components have a

IrinaK [193]

Answer:

following are the solution to this question:

Step-by-step explanation:

In part (a):

Phase 1: Identifies a reasonable confidence period with both the title or form and checks suitable conditions. one z-interval test, p, that true percentage of defective parts in such a lot of Clauses:

1. Unique component sampling

2. This survey size is enough, to the fact to conclude the composition of that percentage of a sample was roughly natural, which the amount for successes and the number of losses are either greater than 10 (or 5 or 15). Needs to check: A 200-component representative selection of a big batch was picked. Each has far more than 15 achievements or failures (16 and 184).

Phase 2: Mechanisms Correct Pick problems for Pink and Blue it Committee of a College. Similar products are part of the program of a College Board. It is prohibited for such resources to be used and distributed online or in a printed form far beyond the participation of the college in the package, which leads to intervals (0.0424, 0.1176).

Phase 3: Background analysis In just this study, we consider 95% of its computed value to have between 4.24% and 11.76% for both the real number of defective components within the batch.

In part (b):

Its standard deviation of 95 percent found in Part (a) allows a company to eliminate class A as its 2 percent faulty rate is not among the computed value within this interval. It period however does not enable the company to decide between levels B and C, since the two faulty rates (5% and 10%) were examined.

Scoring: In 4 components the question is rated. Section 1 comprises part (a) step 1; part (a) or step 2 comprises of the part (a), part (3), part (a), and part (4) (b). The part was marked as correct (E), partly right (P) (I).

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3 years ago

I think its 2, 150 but just making sure.

mina [271]

Answer:

The answer is 2,150

Step-by-step explanation:

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3 years ago

C is the midpoint of AD. B is the midpoint of AC. BC = 12. What is the length of AD

kvasek [131]

48 I would believe is the answer

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3 years ago

Read 2 more answers

(x-4)^2+12(x-4)+20=0

Ratling [72]

Answer:

x=2 or x=−6

Step-by-step explanation:

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3 years ago