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2 years ago

Jennifer painted 168 pots in 14 days. How many pots can she paint in 30 days?

Mathematics

2 answers:

svlad2 [7]2 years ago

6 0

168÷14=12
12×30=360 pots

Vaselesa [24]2 years ago

4 0

The answer would be 360 because if you divide 168 by 14, you'll get the number of pots made each day. And so if 168 pots are made in 14 days, just multiply that by two to get 336 pots made in 28 days. Now, just multiply the number of pots made each day(12) by 2 since there are two remaining days left to get 24. Finally, you add 24 to 336 to get your solution: 360.

168*2
336 pots made in 28 days

12*2
24 pots made in the remaining 2 days left

336+24
360 pots are made in a total of 30 days

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Answer:

0,12 grams

Step-by-step explanation:

0,96:8=0,12

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2 years ago

in september 2016, the cost of gasoline in Berlin was around 1.25 euros per liter. what would the equivalent cost be in u.s. dol

AnnZ [28]

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roughly 1.50

Step-by-step explanation:

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2 years ago

Read 2 more answers

A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t

Alina [70]

Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A bank gets an average of 12 customers per hour.

This means that $\mu = 12$

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

$P(X < 10) + P(X \geq 10) = 1$

We want $P(X \geq 10)$

Then

$P(X \geq 10) = 1 - P(X < 10)$

In which

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-12}*12^{0}}{(0)!} \approx 0$

$P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001$

$P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018$

$P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053$

$P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127$

$P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255$

$P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437$

$P(X = 8) = \frac{e^{-12}*12^{8}}{(8)!} = 0.0655$

$P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874$

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0 + 0.0001 + 0.0004 + 0.0018 + 0.0053 + 0.0127 + 0.0255 + 0.0437 + 0.0655 + 0.0874 = 0.2424$

Then

$P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576$

75.76% probability that there will be 10 or more customers at this bank in one hour.

3 0

2 years ago

Accident rate data y1, ...., y12 were collected over 12 consecutive years t=1,2,...12. At over 12 consecutive years t = 1,2,...,

e-lub [12.9K]

Answer:

The correct option is;

The accident rate is a linear model function of t. After t = 7, the slope changes, with the two lines intersecting at t = 7

Step-by-step explanation:

The given parameters are;

Accident rate data = y₁, y₂, y₃, y₄, y₅, y₆, y₇, y₈, y₉, y₁₀, y₁₁, y₁₂

Time at which data was recorded = t₁, t₂, t₃, t₄, t₅, t₆, t₇, t₈, t₉, t₁₀, t₁₁, t₁₂

Accident rate equation is a linear model given as follows;

y = X·B + E

Where:

y = Accident rate

X = Slope of linear model

B = Year

E = y intercept of model

At the end of the 6th year, a change in a regulation that affects safety, hence accident rate occurred given as follows;

Before the change in safety regulations occurred for year t < 7 y₁ = X₁B + E₁

After the change in safety regulations occurred for year t < 7 y₂ = X₂B + E₂

Therefore the slope changes from X₁ to X₂ after t = 7 with the second linear model starting from the end of the first linear model making the two lines intersect at t = 7 (the beginning of year 7)

Hence the correct option is that "The accident rate is a linear model function of t. After t = 7, the slope changes, with the two lines intersecting at t = 7."

4 0

2 years ago

A farmer estimates that he has 9,000 bees producing honey on his farm. The farmer becomes concerned when he realizes the populat

vivado [14]

Analysis to obtain the function that models the polulaiton ob bees:

1) First year 9,000 bees

2) Second year: decrease 5% => 9,000 - 0.05* 9,000 = 9,000 * (1 - 0.05) = 9,000 * 0.95

3) Every year the population decreases 5% => 9,000 * 0.95)^ (number of years)

4) if you call x the number of years, and f(x) the function that represents the number of bees, then: f(x) = 9,000 (0.95)^ x.

Analysis of the statements:

<span>1) The function f(x) = 9,000(1.05)x represents the situation.

FALSE: WE DETERMINED IT IS f(x) = 9,000 (0.95)^x

2) The function f(x) = 9,000(0.95)x represents the situation.

TRUE: THAT IS WHAT WE OBTAINED AS CONCLUSION OF THE PREVIOUS ANALYSIS.

3) After 2 years, the farmer can estimate that there will be about 8,120 bees remaining.

Do the math:

f(2) = 9,000 * (0.95)^2 = 9,000 * 0,9025 = 8,122

So, the statement is TRUE

4) After 4 years, the farmer can estimate that there will be about 1,800 bees remaining.

f(4) = 9,000 * (0.95)^4 = 9,000 * 0.81450625 = 7,330

So, the statement is FALSE

5) The domain values, in the context of the situation, are limited to whole numbers.

FALSE: THE DOMAIN VALUES ARE ALL NON NEGATIVE REAL VALUES. FOR EXAMPLE THE FUNCTION IS WELL DEFINED FOR X = 5 AND HALF

6) The range values, in the context of the situation, are limited to whole numbers.

TRUE: THERE CANNOT BE FRACTIONS OF BEES
</span>

3 0

2 years ago

Read 2 more answers