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xxMikexx [17]
2 years ago
12

Please help me urgently! 50 points giving brainliest.

Mathematics
1 answer:
Alisiya [41]2 years ago
3 0

The number of significant figures in the numbers are

  • 100 cm = 1 significant digit
  • 0.006700 cm = 2 significant digits
  • 450. cm = 3 significant digits

<h3>How to determine the number of significant figures in the following numbers?</h3>

As a general rule, the zeros before and after the non-zero figures are not significant.

Using the above rule, we have:

  • 100 cm = 1 significant digit
  • 0.006700 cm = 2 significant digits
  • 450. cm = 3 significant digits

<h3>How to round the following numbers to the correct number of significant figures</h3>

Using the above rule in (a), we have:

  • 123g to show 1 sig fig = 100 g
  • 0.19851m to show 2 sig figs = 0.2 m
  • 0.0057034L to show 3 sig figs = 0.005703 L

<h3>How to report the following answers with correct significant figures</h3>

We have:

(12.93cm) x (2.34cm) x (8cm) = 242.05 cm^3 because 12.93 has 4 significant figures

67.0m / 2.18s = 30.7 m/s because 2.18 has 3 significant figures

<h3>How to convert the following metric to metric</h3>

450mL = 0.45 L

because 1 mL = 0.001 L

2.3 dm = 0.00023 km

because 1 dm = 0.0001 km

0.120cg = 1.2 mg

because 1 cg = 10 mg

6700L = 670000 cL

because 1 L = 100 cL

<h3>How to convert the following metric</h3>

a. 2.34miles = 3.76 km (1mile = 1.61km) -- given

b. 5.3ft = 161.544 cm(2.54cm = 1 in)

Because 1 ft = 30.48 cm

Read more about significant figures at:

brainly.com/question/24491627

#SPJ1

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Answer:

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Step-by-step explanation:

you want to set the fractions equal in the denominators:

4/5 x 3= 12/15

you end up with the ratio between 12/15 and 7/15 and you'd get 12 to 7

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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}

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<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

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