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Vikentia [17]
4 years ago
12

a seafood company sold 9,125 pounds of fish last month. if 6 seafood companies sold the same amount of fish how much fish did th

e 6 companies sell last month in all
Mathematics
2 answers:
irina [24]4 years ago
7 0

Answer: There are 54,750 fish that the six companies sold last month in all.

Step-by-step explanation:

Since we have given that

Number of pounds of fish last month = 9,125

Number of sea food companies = 6

So, we need to find the total number of fish that 6 companies sold last month in all.

so, the total number of fish is given by

9125\times 6\\\\=54,750

Hence, there are 54,750 fish that the six companies sold last month in all.

sashaice [31]4 years ago
6 0
Times 9,125 by 6 and you should get 54,750 in total
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Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers have maintained careful records
madreJ [45]

Answer:

For this case we have the following info related to the time to prepare a return

\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation would be:

\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2

And the best answer would be

b. 2 minutes

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we have the following info related to the time to prepare a return

\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation would be:

\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2

And the best answer would be

b. 2 minutes

3 0
3 years ago
How many different arrangements of 5 letters can be made from the 26 letters of the alphabet?
VikaD [51]
Answer: P(26,5).

That is all the possible combinations of 5 letters made out from 26 letters.

That is 26! / [ 5! ( 26 - 5)! ] = 26! / [5! 21! ] = 26*25*24*23*22*21! / [ 5! 21!] =

= 26*25*24*23*22 / [5*4*3*2*1] = 65,780 arrangements.


5 0
4 years ago
Solve the equation 3^(2x) -4*3^(x+1) + 27 = 0. use rules of exponents to change 3^(x+1)​
Leviafan [203]

Answer:

\huge\boxed{x=1\ \vee\ x=2}

Step-by-step explanation:

3^{2x}-4\cdot3^{x+1}+27=0\\\\\text{use}\ (a^n)^m=a^{nm}\ \text{and}\ a^n\cdot a^m=a^{n+m}\\\\\left(3^x\right)^2-4\cdot3^x\cdot3^1+27=0\\\\\left(3^x\right)^2-12\cdot3^x+27=0\\\\\text{substitute}\ 3^x=t>0\\\\t^2-12t+27=0\\\\t^2-3t-9t+27=0\\\\t(t-3)-9(t-3)=0\\\\(t-3)(t-9)+0\iff t-3=0\ \vee\ t-9=0\\\\t-3=0\qquad\text{add 3 to both sides}\\\boxed{t=3}\\\\t-9=0\qquad\text{add 9 to both sides}\\\boxed{t=9}

\text{We return to substitution:}\\\\3^x=t\\\\3^x=3\ \vee\ 3^x=9\\\\3^x=3^1\ \vee\ 3^x=3^2\\\\\boxed{x=1}\ \vee\ \boxed{x=2}

4 0
3 years ago
Los puntajes obtenidos por un grupo de alumnos en un examen de 10 puntos son 3,5,2,6,5,9,5,2,8,6. Para saber el puntaje máximo q
Vikentia [17]

Answer:

Para saber el puntaje máximo que obtuvo el 75% de los alumnos se debe calcular el tercer cuartil.

Step-by-step explanation:

Los Cuantiles (cuartiles, deciles, percentiles) son medidas de localización cuya  función es informar del valor de la variable que ocupará la posición (en tanto por  cien) que interese respecto de todo el conjunto de variables.

Los cuartiles son medidas estadísticas de posición que tienen la  propiedad de dividir la serie estadística en cuatro grupos de números iguales de  términos.

El primer cuartil o cuartil inferior es aquel valor de la variable tal que la cuarta parte (25%) de las observaciones son inferiores o iguales a él, y el resto (75%) es superior o igual. El segundo cuartil es la mediana, ya que se trata del valor localizado en la mitad de la distribución. Finalmente, el tercer cuartil o cuartil superior es un valor tal que las tres cuartas partes de las observaciones son inferiores o iguales a él.

En otras palabras, el primer cuartil Q1, es el valor en el cual o por debajo del cual queda aproximadamente un cuarto (25%) de todos los valores de la sucesión (ordenada); El segundo cuartil Q2 es el valor por debajo del cual queda el 50% de los datos (Mediana), el tercer cuartil Q3 es el valor por debajo del cual quedan las tres cuartas partes (75%) de los datos. Es decir, Q1, Q2 y Q3 determinan los valores correspondientes al 25%, al 50% y al 75% de los datos respectivamente. Q2 coincide con la mediana.

<u><em>Para saber el puntaje máximo que obtuvo el 75% de los alumnos se debe calcular el tercer cuartil.</em></u>

<u><em></em></u>

5 0
3 years ago
Simplify the expression: 4(10 + 6u)
Marianna [84]

Answer:

24u + 10

Step-by-step explanation:

first rearrange the terms

4(10+6u)

4(6u+10)

then distribute the 4

4(6u) = 24u

4(10) = 10

then you get 24u + 10

7 0
3 years ago
Read 2 more answers
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