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2 years ago

A bag can hold 1 1/2pounds of flour. if Mimi has 7 1/2pounds of flour, then how many bags of flour can Mimi make?

Mathematics

2 answers:

Naily [24]2 years ago

6 0

Answer:

5 BAGS OF FLOUR

7 1/2 ÷ 1 1/2 = 5

Based on the given conditions, formulate: 7 1/2 / 1 1/2 Find common denominator and write the numeral.

umka21 [38]2 years ago

4 0

Answer:

7 1/2 ÷ 1 1/2

make them improper fractions;

15/2 ÷ 3/2

make 3/2 a reciprocal(basically flipping the fraction upside down)

15/2 ÷ 2/3

and multiply across;

15/2 x 2/3 = 30/6 which simplifies to 5 bags of flour.

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A new pain reliever is advertised as being 97%

Likurg_2 [28]

Answer:

B. 0.694

Step-by-step explanation:

Probability is the ratio of the number of possible outcome to the number of total outcome.

Given that the pain reliever is advertised as being 97% effective, for it to be effective in 12 independent trials

= (97%) ^ 12

= 0.693842361

The is approximately 0.694 to 3 decimal places

6 0

3 years ago

Read 2 more answers

If an event has a 55% chance of happening in one trial, how do I determine the chances of it happening more than once in 4 trial

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The chances of it happening more than once in 4 trials is 13%

<h3>How to determine the number</h3>

From the information given, we have can deduce that;

Probability of 1 trial = 55%

= 55/ 100

Find the ratio

= 0. 55

We are to find the probability of it happening more than once in 4 different trials

If the probability of it happening in one trial is 555 which equals 0. 55

Then the probability of it happening in 1 in 4 trials is given as;

P(1/4 trials) = 1/ 4 × 55%

P(1/4 trials) = 1/ 4 × 0. 55

Put in decimal form

P(1/4 trials) = 0. 25 × 0. 55

P(1/4 trials) = 0. 138

But we have to know the percentage

= 0. 138 × 100

Multiply the values, we have

= 13. 8 %

Thus, the chances of it happening more than once in 4 trials is 13%

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2 years ago

Anyone please gives me the correct answer to this problem on Khan Academy. The assignment is due tomorrow, and I don't want to f

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Answer:

(-2,-2) and (-7,1)

Step-by-step explanation:

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2 years ago

Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.

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Let $z=-\sqrt3+i$ . Then

$|z|=\sqrt{(-\sqrt3)^2+1^2}=2$

$z$ lies in the second quadrant, so

$\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6$

So we have

$z=2e^{i5\pi/6}$

and the fourth roots of $z$ are

$2^{1/4}e^{i(5\pi/6+k\pi)/4}$

where $k\in\{0,1,2,3\}$ . In particular, they are

$2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}$

$2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}$

$2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}$

$2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}$

7 0

3 years ago

Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all

posledela

This question is Incomplete

Complete Question

Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all have the same body size in their adult phase, which made it easy to measure speeds in units of body lengths per second (bl/s). The researchers found that, when traffic is light and not congested, ant speeds vary roughly Normally, with mean 6.20 bl/s and standard deviation 1.58 bl/s. (a) What is the probability that an ant's speed in light traffic is faster than 5 bl/s? You may find Table B useful. (Enter your answer rounded to four decimal places.)

Answer:

0.7762

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Population mean = 6.20 bl/s

Standard deviation = 1.58 bl/s.

x = 5 bl/s

z = 5 - 6.20/1.58

z = -0.75949

The probability that an ant's speed in light traffic is faster than 5 bl/s is P( x > 5)

Probability value from Z-Table:

P(x<5) = 0.22378

P(x>5) = 1 - P(x<5)

= 1 - 22378

= 0.77622

Approximately to 4 decimal places = 0.7762

The probability that an ant's speed in light traffic is faster than 5 bl/s is 0.7762

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3 years ago