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2 years ago

8

what is the smallest number of integer kilogram weights needed to weigh every whole kilogram between 1kg and 40kg?

1 answer:

lisabon 2012 [21]2 years ago

6 0

9514 1404 393

Answer:

4

Step-by-step explanation:

We assume you are using a balance, and that weights can be put in either pan. In that case, weight values can be either added or subtracted, so there are 3 possibilities for the contribution each weight makes. In other words, we need to find how many digits are required to count to 40 in base 3.

The largest 3-digit number in base 3 is ...

3^2 +3^1 + 3^0 = 13 (base 10)

The largest 4-digit number adds 3^3 = 27 to this: 27 +13 = 40.

Hence, we can weigh values 0 to 40 using 4 weights, of sizes 1, 3, 9, 27.

__

Some example weights are ...

5 = 9 - 3 - 1

38 = 27 + 9 + 3 - 1

where the "negative" weights are placed in the same pan as the unknown.

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A study found that, in 2005, 12.5% of U.S. workers belonged to unions (The Wall Street Journal, January 21, 2006). Suppose a sam

Julli [10]

Answer:

There is not enough evidence to support the claim that union membership increased.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 400

p = 12.5% = 0.125

Alpha, α = 0.05

Number of women belonging to union , x = 52

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.125\\H_A: p > 0.125$

The null hypothesis sates that 12.5% of U.S. workers belong to union and the alternate hypothesis states that there is a increase in union membership.

This is a one-tailed(right) test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{52}{400} = 0.13$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.13-0.125}{\sqrt{\frac{0.125(1-0.125)}{400}}} = 0.3023$

Now, we calculate the p-value from the table.

P-value = 0.3812

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Conclusion:

Thus, there is not enough evidence to support the claim that union membership increased.

8 0

2 years ago

Pls can someone help me and have it right

dangina [55]

Answer:

Problem 1:

a. x=2

b. x=3

c. x=1

Problem 2:

A multiplication equation to hold the table true:

$18\times3=54$

A division equation to hold the table true:

$\frac{54}{3} =18$

Step-by-step explanation:

Given in problem 1:

(a). The equation is $\frac{x}{6} > 1$

It holds true for all values of $x> 6$ .

Let us say $x=12$ ,

$\frac{x}{6}=\frac{12}{6} =2$ which is greater than 1.

(b). The equation is $\frac{x}{6} < 1$

It holds true for all values of $x< 6$ .

Let us say $x=3$

$\frac{x}{6} =\frac{3}{6} =\frac{1}{2}$ which is less than 1.

(c). The equation is $\frac{x}{6} = 1$

It holds true for only $x= 6$ .

Let us say $x=6$ ,

$\frac{x}{6} =\frac{6}{6}=1$ which is equal to 1.

Problem 2:

A multiplication equation to hold the table true:

$18\times3=54$

A division equation to hold the table true:

$\frac{54}{3} =18$

Therefore these are the values which hold true to the equation in problem 1 and 2.

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3 years ago

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Answer:

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Step-by-step explanation:

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Margarita [4]

Answer:

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Step-by-step explanation:

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Maksim231197 [3]

250 • .15 = 37.50
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