Arrange the following<br> weights in descending<br> order<br> 50 mg, 35 hg, 500g, 14 KG

Can someone help me with this problem!

Juli2301 [7.4K]

-5 I think

Ignore this this is just for 20 words

3 0

2 years ago

Compare the dotplot to a histogram of the data. (Select all that apply.)

Aleks04 [339]

Answer:

(B) The raw data can be retrieved from the dot plot, but not the histogram.

(C) Dot plots show the frequency of individual data values.

Step-by-step explanation:

There are two correct answers:

(B) The raw data can be retrieved from the dot plot, but not the histogram.

(C) Dot plots show the frequency of individual data values.

Dot plots are used for small data sets where values fall into a number of categories, unlike histograms.

8 0

2 years ago

The table shows the number of miles that a car travels and the gallons of gasoline it uses to go those distances.

PtichkaEL [24]

Answer:

4 -.-

Step-by-step explanation:

8 0

2 years ago

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

2 years ago

You spin the spinner, flip a coin, then spin the spinner again. Find the probability of the compound event. Write your answer as

Lesechka [4]

The correct question is;

You spin the spinner, flip a coin, then spin the spinner again. Find the probability of the compound event. Write your answer as a fraction or percent. If necessary round your answer to the nearest hundredth. Find the probability of spinning an odd number, flipping heads, then spinning another odd number? Write your answer as a fraction in simplest form.

Answer:

If the spinner has an even number of spaces, the probability of spinning an odd number, flipping heads, then spinning a yellow is = 1/8

Whereas, if the spinner has an odd number of spaces, more information is needed

Step-by-step explanation:

The answer will depend on whether the spinner has an odd or even number of spaces.

Now, if we suppose that the spinner has an even amount of numbers on it, there will be an equal number of odd and even numbers.

Thus the probability of spinning an odd number is ½.

Also, the probability of flipping heads on a coin is ½.

Thus, the probability of spinning an odd number, flipping heads, then spinning another odd number is;

= ½ × ½ × ½ = ⅛

Similarly,

If the spinner has an odd amount of numbers on it, the probability of spinning an odd number will vary. For instance, if there are 5 spaces on the spinner, the probability of spinning an odd number will be

3

However, if there are 9 spaces, the probability will be 5/9

So,we can conclude that;

If the spinner has an even number of spaces, the probability of spinning an odd number, flipping heads, then spinning a yellow is = 1/8

Whereas, if the spinner has an odd number of spaces, more information is needed.

6 0

2 years ago

Arrange the following weights in descending order 50 mg, 35 hg, 500g, 14 KG