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3 years ago

Workers have packed 1,400 glasses in 7 boxes. If they pack 3 more boxes, how many glasses will they have packed in all?

Mathematics

2 answers:

11111nata11111 [884]3 years ago

8 0

1box = 1400/7 = 200

200×3=600

1400+600=2000

sdas [7]3 years ago

4 0

Answer:

2000

Step-by-step explanation:

Given :Workers have packed 1,400 glasses in 7 boxes.

To Find :If they pack 3 more boxes, how many glasses will they have packed in all?

Solution:

Workers packed no. of glasses in 7 boxes = 1400

Workers packed no. of glasses in 1 box = $\frac{1400}{7}$

Workers packed no. of glasses in 3 boxes = $\frac{1400}{7} \times 3$

= $600$

So, initially they packed 1400 glasses

If they pack 3 more boxes so, the pack 600 glasses more

So, The total no. of glasses have packed by workers = 1400+600 = 2000

Hence they have packed 2000 glasses in all.

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kogti [31]

It is good equation. Look, that:
81 = 3 x 3 x 3 x 3 = 3^4.
So you have to solve:
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exponents have to be the same, so
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3 years ago

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Kipish [7]

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Step-by-step explanation:

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3 years ago

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home. Give the value of th

astra-53 [7]

Answer:

The value of the standard error for the point estimate is of 0.0392.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home.

This means that $n = 100, p = \frac{81}{100} = 0.81$

Give the value of the standard error for the point estimate.

This is s. So

$s = \sqrt{\frac{0.81*0.19}{100}} = 0.0392$

The value of the standard error for the point estimate is of 0.0392.

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3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

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3 years ago

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raketka [301]

Answer:

$f(x)=\frac{1}{x-7}-3$ , see the graph attached for a visual reference.

Step-by-step explanation:

Vertical asymptotes are only present in rational functions where the parent function $f(x)=\frac{1}{x}$ has a vertical asymptote at the line $x=0$ and a horizontal asymptote at the line $y=0$ . Because the vertical asymptote has to be $x=7$ , the denominator must be x-7 in order for the denominator to equal 0. For the horizontal asymptote to be $y=-3$ , then 3 must be subtracted from the rational function. Therefore, the function that has these asymptotes is $f(x)=\frac{1}{x-7}-3$ .

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3 years ago

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