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4 years ago

LOTS OF POINTS!! 1ST GRADE QUESTION!! MY CALCULATOR ISNT WORKING!!!

Mathematics

2 answers:

MArishka [77]4 years ago

6 0

Answer:

No, √3 is not rational. If you need more of an explanation let me know

Mashutka [201]4 years ago

6 0

just look on Google!

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Pete and his friend Danny volunteered to bring sandwiches for lunch at a homeless shelter. They bought 7 loaves of bread. Each l

Gnesinka [82]

Answer:

63 sandwiches

Step-by-step explanation:

Pete and his friend Danny volunteered to bring sandwiches for lunch at a homeless shelter

They brought 7 loaves of bread

Each loaves have 18 slices

They used 2 slices each to make the sandwich

The first step is to calculate the total number of slices

Sinces 7 loaves of bread are used and each of them contain 18 slices then, the total number of slices can be calculated as follows

= 7 ×18

= 126 slices

Therefore since 2 slices of bread is used to make one sandwich then, the number of sandwiches that was made can be calculated as follows

= 126/2

= 63

Hence 63 sandwiches were made for the shelter

8 0

3 years ago

Mr. Clifton was preparing for his upcoming birthday party. He ordered 51 white balloons and 47 red balloons. As he walked to his

Andrej [43]

Answer:

28 i think

Step-by-step explanation:

51+47=98

98-12 the ones that flew away that takes it to 86

86/3 the 3 balloons buntch than is 28

p.s. i mght of got some equations wrong but this is what i got

4 0

3 years ago

Two sides of an equilateral triangle, ABC, are AB = 2x – 2 and BC = 32 – 6.

IrinaVladis [17]

Here yah go hope this is it

8 0

3 years ago

Please help <br> What is the slope of a line that is parallel to the line shown?

Jlenok [28]

Answer:

m=2/3

Step-by-step explanation:

your answer would be A.

8 0

3 years ago

Read 2 more answers

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago