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1 year ago

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A bakery sold a total of 3028 coffee buns and blueberry buns. 1560 more coffee buns were sold than the blueberry buns. How many

coffee buns did the bakery sell?
P.S. No algebra to be used here as it is a grade 4th question!

1 answer:

Natali [406]1 year ago

7 0

Answer:

$\large{ \tt{ - \: HEY \: AH~\:♡ }}$

$\large{ \tt{✺ \: SOLUTION}} :$

Provided : Total sold bakery items : 3028 coffee buns and blueberry buns & 1560 more coffee buns were sold than the blueberry buns.

To find : Number of coffee buns the bakery sold

- First , Subtract 1560 from 3028 :

$\large{ \tt{→ \: 3028 - 1560 = 1468}}$

We just subtacted the number of more coffee buns from the number of told items sold which means that the number of coffee buns and the number of blueberry buns sold are equal for now. Now divide 1428 by 2 :

$\large{ \tt{→ \frac{1428}{2} = 734 }}$

Now - Let's get back to the second sentence of the question and add 734 & 1560 :

$\large{ \tt{→ \: 734 + 1560 = \large{ \boxed{2294}}}}$

Hence , The bakery sold 2294 coffee buns.

- Hope this helps , oneesan! ;)

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The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

1 year ago

Mr. mole left his burrow that lies 7 meters below the ground and started digging his way deeper into the ground, descending at a

Keith_Richards [23]

Answer:

A(t)=−1.8t−4.5

Step-by-step explanation:

Mr. Mole is descending at a constant rate, so we are dealing with a linear relationship.

We know that Mr. Mole descended at a rate of 1.81.81, point, 8 meters per minute, so the slope \green mmstart color green, m, end color green is \green{-1.8}−1.8start color green, minus, 1, point, 8, end color green, and our function looks like A(t)=\green{-1.8}t+\pink bA(t)=−1.8t+bA, left parenthesis, t, right parenthesis, equals, start color green, minus, 1, point, 8, end color green, t, plus, start color pink, b, end color pink.

We also know that after 555 minutes, Mr. Mole was 13.513.513, point, 5 meters below the ground, which means that A(5)=-13.5A(5)=−13.5A, left parenthesis, 5, right parenthesis, equals, minus, 13, point, 5. We can substitute this into the formula of the function to find \pink bbstart color pink, b, end color pink:

\qquad\begin{aligned}A(5)&=-13.5\\ \green{-1.8}\cdot5+\pink b&=-13.5\\ -9+\pink b&=-13.5\\ \pink b&=\pink{-4.5}\end{aligned}

A(5)

−1.8⋅5+b

−9+b

b

=−13.5

=−13.5

=−13.5

=−4.5

This means that Mr. Mole's initial altitude is 4.54.54, point, 5 meters below the ground.

AKA the answer is A(t)=−1.8t−4.5

3 0

2 years ago

Two commuters leave the same city at the same time but travel in opposite directions. One car is traveling at an average speed o

r-ruslan [8.4K]

Answer:

5 hours

Step-by-step explanation:

The formula to relate speed, distance and time is:

D = V * t

Where D is the distance, V is the speed and t is the time.

As they are in opposite directions, we need to sum the speed.

So the total speed is:

V = 63 + 59 = 122 mph

So using this value and V and using D = 610, we have:

610 = 122 * t

t = 610 / 122 = 5 hours

So they will be 610 miles apart after 5 hours.

3 0

2 years ago

Zucchini plants require 9 square feet around each plant.

MrRa [10]

The area of square TUVW = 3×3 = 9 square units

1 square unit = 4 head lettuces
9 square units = 9×4 = 36 lettuces

Area of plot QRST = length × width = 12×6 = 72 square units

36 vegetables for every 9 square units
We have 72÷9 = 8 lots of 9 square units

Total vegetable = 8 × 36 = 288 vegetables

We can have 8 different vegetables, each type can have 36 vegetables

4 0

1 year ago

The statement tan theta -12/5, csc theta -13/5, and the terminal point determined by theta is in quadrant 2."

Harlamova29_29 [7]

Answer:

Answer C:

Cannot be true because $csc(\theta)$ is greater than zero in quadrant 2.

Step-by-step explanation:

When the csc of an angle is negative, since the cosecant function is defined as:

$csc(\theta)=\frac{1}{sin(\theta)}$

that means that the sin of the angle must be negative, and such cannot happen in the second quadrant. The sine function is positive in the first and second quadrant.

Therefore, the correct answer is:

Cannot be true because $csc(\theta)$ is greater than zero in quadrant 2.

6 0

2 years ago