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3 years ago

A brick layer is able to set 2.5 breaks in one minute how many bricks can you set an eight hours

Mathematics

1 answer:

frozen [14]3 years ago

5 0

Answer:

<h2> In 8 hours he will set 1200 bricks</h2>

Step-by-step explanation:

In this problem, we are expected to estimate the number of bricks he can set in 8 hours.

let us convert hours to minutes we have

60 min make 1 hour

x min make 8 hours

cross multiply

x= 60*8

x= 480 min

We are assuming the same rate with which he sets 2.5 brick is the same as the rate with which he will set bricks for the 8 hours windows

so if the bricklayer sets

2.5 bricks in 1min

y bricks in 480 min

Cross multiply we have

y=2.5*480

y= 1200 bricks

hence in 8 hours, he will set 1200 bricks

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4. Find the area of the triangle whose breadth is 48 cm and height is 5cm

ryzh [129]

Answer:120

Step-by-step explanation:

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3 years ago

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A manufacturer of processing chips knows that 2\%2%2, percent of its chips are defective in some way. Suppose an inspector rando

kipiarov [429]

The data in the question seems a bit erroneous. I am writing the correct question below:

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it.

Answer:

The probability that at least one of the selected chips is defective is 0.0776.

Step-by-step explanation:

The question states that the probability of defective chips is 2% i.e. 0.02. Let p denote the probability of selecting a defective chip so, p = 0.02

An inspector selects 4 chips, which means n=4 and we need to compute the probability that at least one of the selected chips is defective. Let X be the number of defective chips selected. We need to compute P(X≥1) which means either 1, 2, 3 or 4 chips can be defective.

We will use the binomial distribution formula to solve this problem. The formula is:

where n = total no. of trials

p = probability of success

x = no. of successful trials

q = probability of failure = 1-p

we have n=4, p=0.02 and q=1-0.02=0.98.

We need to compute P(X≥1) which is equal to:

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

A shorter method to do this is to use the total probability theorem:

P(X≥1) = 1 - P(X<1)

= 1 - P(X=0)

= 1 - ⁴C₀ (0.02)⁰(0.98)⁴⁻⁰

= 1 - (0.98)⁴

= 1 - 0.9224

P(X≥1) = 0.0776

4 0

4 years ago

A coin is tossed and a single 6 sided number cube is rolled. Find the probability of landing on the heads side of the coin and r

Delicious77 [7]

Answer:

The probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

Step-by-step explanation:

Given:

Coin is tossed and a single 6 sided number cube is rolled.

Let the event of tossing head be 'H' and event of rolling a 3 be 'R3'.

Now, we know that:

Probability of an event 'A' = Favorable outcomes of 'A' ÷ Total possible outcomes

Now, for tossing a coin, the total possible outcomes are head and tail. So, there are 2 possible outcomes.

So, probability of event 'H' is given as:

$P(H)=\dfrac{n(H)}{n(S)}\\\\P(H)=\frac{1}{2}=0.5$

Similarly, for rolling a 6 sided cube, the possible outcomes are numbers 1 to 6. So, the number of possible outcomes is, $n(S)=6$

Now, probability of rolling a 3 is given as:

$P(R3)=\frac{n(R3)}{n(S)}\\\\P(R3)=\frac{1}{6}$

Now, both the events 'H' and 'R3' occur together. So, the combined probability is the product of two individual probabilities as they are independent events. So,

$P(H\ and\ R3)=P(H)\times P(R3)\\\\P(H\ and\ R3)=\frac{1}{2}\times \frac{1}{6}\\\\P(H\ and\ R3)=\frac{1}{12}=0.083$

Note: Independent events are those events whose intersection is an empty set of events or the outcome of one event doesn't affect the outcome of the other event.

Therefore, the probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

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3 years ago

HELP WILL MARK YOU BRAINLIEST NO GUESSES PLZ!!!!

bazaltina [42]

the answer is D and i know cause i just did that and thats the answer

7 0

3 years ago

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3. Classify each function according to whether it is a vertical stretch, a vertical compression, a horizontal

Vladimir79 [104]

The classifications of the functions are

A vertical stretch --- p(x) = 4f(x)
A vertical compression --- g(x) = 0.65f(x)
A horizontal stretch --- k(x) = f(0.5x)
A horizontal compression --- h(x) = f(14x)

<h3>How to classify each function accordingly?</h3>

The categories of the functions are given as

A vertical stretch
A vertical compression
A horizontal stretch
A horizontal compression

The general rules of the above definitions are:

A vertical stretch --- g(x) = a f(x) if |a| > 1
A vertical compression --- g(x) = a f(x) if 0 < |a| < 1
A horizontal stretch --- g(x) = f(bx) if 0 < |b| < 1
A horizontal compression --- g(x) = f(bx) if |b| > 1

Using the above rules and highlights, we have the classifications of the functions to be

A vertical stretch --- p(x) = 4f(x)
A vertical compression --- g(x) = 0.65f(x)
A horizontal stretch --- k(x) = f(0.5x)
A horizontal compression --- h(x) = f(14x)