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lukranit [14]
3 years ago
8

A board measures 34.1 cm. What is the greatest possible error allowed for the measurement?

Mathematics
1 answer:
gladu [14]3 years ago
7 0
Greatest error is half smallest measuring unit. It's measured to 0.1 cm , half that is 0.05 cm answer D.
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A flower pot weighing 3 newtons is sitting on a windowsill 10 meters from the ground what is its potential energy
krek1111 [17]

Answer:

30 J

Step-by-step explanation:

W = mg = 3 N

h = 10 m

Potential Energy = mgh = 3*10 = 30 J

5 0
3 years ago
5/7+(-1/3)-(-1/2)<br><br> Show work pls
ELEN [110]
30/42-14/42+21/42= 37/42
7 0
3 years ago
Whats 4n = 880.4's solution??
vodka [1.7K]

Answer:

n = 220.1

Step-by-step explanation:

  • 4n = 880.4
  • n = 880.4 ÷ 4
  • n = 220.1

the value of n is 220.1

6 0
2 years ago
I need help asap!!!!!
Svetllana [295]
Answer: 130

In order to solve for <4, you need to know that supplementary angles add up to 180°. 
<3 = 50
180 - 50
130

**For future questions**
1. The sum of the interior angles of a triangle is 180°
2. Alternate interior angles are congruent
3. The exterior angle of a triangle is the sum of the nonadjacent interior angles

See attachment below. 

4 0
3 years ago
A manager is comparing wait times for customers in a coffee shop based on which employee is
anyanavicka [17]

Using the t-distribution, as we have the standard deviation for the sample, it is found that there is a significant difference between the wait times for the two populations.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if there is no difference, that is:

H_0: \mu_A - \mu_B = 0

At the alternative hypothesis, it is tested if there is difference, that is:

H_1: \mu_A - \mu_B = 0

<h3>What are the mean and the standard error of the distribution of differences?</h3>

For each sample, we have that:

\mu_A = 73, s_A = \frac{2}{\sqrt{100}} = 0.2

\mu_B = 74, s_B = \frac{4}{\sqrt{100}} = 0.4

For the distribution of differences, we have that:

\overline{x} = \mu_A - \mu_B = 73 - 74 = -1

s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.2^2 + 0.4^2} = 0.447

<h3>What is the test statistic?</h3>

It is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 0 is the value tested at the null hypothesis.

Hence:

t = \frac{\overline{x} - \mu}{s}

t = \frac{-1 - 0}{0.447}

t = -2.24

<h3>What is the p-value and the decision?</h3>

Considering a one-tailed test, as stated in the exercise, with 100 - 1 = 99 df, using a t-distribution calculator, the p-value is of 0.014.

Since the p-value is less than the significance level of 0.05, it is found that there is a significant difference between the wait times for the two populations.

More can be learned about the t-distribution at brainly.com/question/16313918

8 0
2 years ago
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