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1 year ago

A two pound box of fruit snacks contains 24 packets. Find the unit rate in packets per pound

Mathematics

2 answers:

Monica [59]1 year ago

6 0

24/2=12

The unit rate is 12 packets per pound.

Hope this helps!

Evgesh-ka [11]1 year ago

6 0

Answer:

Step-by-step explanation:

Given that a two pound box of fruit snacks contains 24 packets

We have to find the unit rate per pound

This is a question of direct variation.

$2 pounds = 24 packets\\1 pount= 12 packet$

i.e. unit rate is 12 packets per pound

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Suppose that the ages of members in a large billiards league have a known standard deviation of σ = 12 σ=12sigma, equals, 12 yea

asambeis [7]

Answer:

$n\geq 23$

Step-by-step explanation:

-For a known standard deviation, the sample size for a desired margin of error is calculated using the formula:

$n\geq (\frac{z\sigma}{ME})^2$

Where:

$\sigma$ is the standard deviation
$ME$ is the desired margin of error.

We substitute our given values to calculate the sample size:

$n\geq (\frac{z\sigma}{ME})^2\\\\\geq (\frac{1.96\times 12}{5})^2\\\\\geq 22.13\approx23$

Hence, the smallest desired sample size is 23

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1 year ago

Mario, Yoshi, and Toadette play a game of "nonconformity": They each choose rock, paper, or scissors. If two of the three people

andrey2020 [161]

Step-by-step explanation: Hi!

well, the only case where someone wins is if two of them chose the same thing, and the other chose one of the two remaining possibilities.

an useful way of starting this is calculated al the possible combinations of rock, paper and scissors.

we have 3 persons and 3 options, ence the total number of permutations is 3*3*3 = 27

So, the cases where none wins are:

1) the 3 chose the same thing: there are 3 combinations here, one for each option

2) the 3 chose a different thing: here you think, the first one has 3 possibilities, then the second one must choose a different option, so has 2, and the third one only has 1 remaining option. so you have 6 combinations.

so there are 6 + 3 = 9 combinations where no one wins, if we want to se the probability, we compute 9/27 = 1/3 = 0.33%.

But the problem ask for it to happen 4 times in a row, so you will have 0.33*0.33*0.33*0.33 = 1/81 chances that this happens

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2 years ago

Bill launched a model rocket, and estimated its height h, In feet, after 1 seconds. His results are shown in the table

gizmo_the_mogwai [7]

Answer:

Bill launched a model rocket, and estimated its height h, In feet, after 1 seconds. His results are shown in the table

Time, 1 0 1 2 3 4

Height, h 0 110 190 240 255

Bill's data can be modeled by the function h(t) = -1612 + 128.

Which value is the best prediction for the height of the rocket after 5.5 seconds?

A 150 ft

B. 180 ft

C. 220 ft

D. 250 ft

E 260 ft

0 1

1 year ago

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

2 years ago

Choose the table that represents g(x) = −2⋅f(x) when f(x) = x + 4.

grigory [225]

Answer: THIRD OPTION.

Step-by-step explanation:

Since $g(x) = -2f(x)$ , then $g(x)$ is:

$g(x)=-2( x + 4)\\\\g(x)=-2x-8$

In order to find which table represents $g(x) = -2f(x)$ when $f(x) = x + 4$ , you can follow these steps:

1. Substitute $x=1$ into the function $g(x)=-2x-8$ :

$g(1)=-2(1)-8\\\\g(1)=-10$

2. Substitute $x=2$ into the function $g(x)=-2x-8$ :

$g(2)=-2(2)-8\\\\g(2)=-12$

3. Substitute $x=3$ into the function $g(x)=-2x-8$ :

$g(3)=-2(3)-8\\\\g(3)=-14$

Notice that those ouput values (for $x=1,\ x=2$ and $x=3$ ) matches with the table provided in the third option. That table represents $g(x) = -2f(x)$ when $f(x) = x + 4$ :

6 0

1 year ago

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