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Juliette [100K]
3 years ago
6

A rucksack can hold a maximum of 6 kg.

Mathematics
2 answers:
Lemur [1.5K]3 years ago
7 0
(convert the 6kg to g first)

1kg = 1000
6kg=6000

6000 ÷ 680 = 8,82

the rucksack can hold between 8 and 9 textbooks
ratelena [41]3 years ago
5 0

Answer:

9 textbooks!

Step-by-step explanation:

Rucksack can hold = 6kg

Mass of textbook = 680g

So, here we need to show the division for finding the number of textbooks it can hold.

680 g in KG's = 0.68

Now, 6/0.68

That is 8.82 ( Round off = 9 textbooks)

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Complementary events must have a sum of 1.
Bas_tet [7]

This is a true statement.

Consider two events A and B. We say they are complementary if P(A)+P(B) = 1

This means that either event A or event B must happen, since the "1" represents 100% probability. Having a probability of 100% means absolute certainty.

---------

Example:

A = the event it rains

B = the event it does not rain

P(A) = 0.30 = 30% chance of rain

P(B) = 0.70 = 70% chance it does not rain

P(A)+P(B) = 0.30+0.70 = 1

So this shows the two events are complementary.

8 0
4 years ago
Consider independent simple random samples that are taken to test the difference between the means of two populations. The varia
Arturiano [62]

Answer:

d. t distribution with df = 80

Step-by-step explanation:

Assuming this problem:

Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:

a. t distribution with df = 82.

b. t distribution with df = 81.

c. t distribution with df = 41.

d. t distribution with df = 80

Solution to the problem

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

This last one is an unbiased estimator of the common variance \sigma^2

So on this case the degrees of freedom are given by:

df= 37+45-2=80

And the best answer is:

d. t distribution with df = 80

5 0
3 years ago
You are competing in a race. The table shows the times from last year's race. You want your time to be last year's median time w
olasank [31]

Answer:

  28 ≤  x ≤40

Step-by-step explanation:

Arrange the data from the smallest to the largest

26, 27, 28,32, 32, 33, 35, 36, 38, 40,42,48

Identify the median value

26, 27, 28,32, 32,<u> 33, 35</u>, 36, 38, 40,42,48

Getting the median

(33+35)/2   =  68/2  = 34 minutes

You want your time to be last year's median time with an absolute deviation of at most 6 minutes

This mean ±6 minutes, hence this will be 34 ± 6

34+6=40  and 34-6=28

The inequality will be    28 ≤  x ≤40

7 0
3 years ago
Read 2 more answers
Help is appreciated ty!!
Ket [755]

Answer:

B and C

Step-by-step explanation:

Jane found multiples of 8 instead of factors and should've gotten 1, 2, 4, and 8 as her answer.

Hope this helps :)

5 0
2 years ago
Read 2 more answers
Please help me on this !!
aev [14]

Answer:

<h3>man tht is hard <em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>mhmm</u></em></h3>

Step-by-step explanation:

srry but can't jelp

6 0
3 years ago
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