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3 years ago

PLEASE HURRY

Mathematics

1 answer:

mina [271]3 years ago

5 0

Answer:

X ≥ 3

Step-by-step explanation:

Let the number of grapes you'll buy be X, you need to buy at leaF 3 pounds so:

X ≥ 3

Answered by Gauthmath

You might be interested in

Solve 2x^2=11x+6 by factoring

Ede4ka [16]

Answer:

(x-6) (2x+1)

Step-by-step explanation:

1) move everything over to the left side, so subtract 11x and 6.

2x^2 -11x - 6 = 0

2)multiply 2(a term) with -6 (c term)

x^2 -11x -12

3) factor and find what multiples to -12 and adds up to -11. in this case its -12 and positive 1

(x-12) (x+1)

4) divide -12 and 1 by the original a term (2)

(x-6) (x+1/2)

5) move the denominator of 2 over to the x.

(x-6) (2x+1)

5 0

3 years ago

La casa de Mauricio está construida en un terreno de forma rectangular, el cual mide el doble de largo que de ancho. Mauricio as

elixir [45]

Answer:

x^2 = 121

Step-by-step explanation:

Sea el ancho del terreno x metros

Dado que la longitud es el doble del ancho, entonces la longitud será 2 * x = 2x

Matemáticamente, el área de un rectángulo es;

Anchura longitud Por lo tanto, tenemos

x * 2x = 242 metros cuadrados

Así; 2x ^ 2 = 242

x ^ 2 = 242/2

x ^ 2 = 121

7 0

2 years ago

What is 3/4 + 3.3 - 1 1/2?

guapka [62]

Answer:

2.55

Step-by-step explanation:1

Convert 1121\frac{1}{2}121 to improper fraction. Use this rule: abc=ac+bca \frac{b}{c}=\frac{ac+b}{c}acb=cac+b.

34+3.3−1×2+12\frac{3}{4}+3.3-\frac{1\times 2+1}{2}43+3.3−21×2+1

Simplify 1×21\times 21×2 to 222.

34+3.3−2+12\frac{3}{4}+3.3-\frac{2+1}{2}43+3.3−22+1

Simplify 2+12+12+1 to 333.

34+3.3−32\frac{3}{4}+3.3-\frac{3}{2}43+3.3−23

Simplify.

5.12\frac{5.1}{2}25.1

Simplify.

2.552.552.55

6 0

2 years ago

44 is what percent of 52?<br> please help

Andrews [41]

Answer:

44% of 52

44/100 ×52

22.88

4 0

3 years ago

A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January

Blababa [14]

Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

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.

23% of the respondents have boycotted goods for ethical reasons.

This means that $p = 0.23$

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

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

$P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789$

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084$

$P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181$

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

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

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439$

$P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109$

$P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116$

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

$E(X) = np = 10*0.23 = 2.3$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33$

The expected number is 2.3 and the standard deviation is 1.33.

5 0

3 years ago