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3 years ago

What is true about days and min.

Mathematics

1 answer:

pav-90 [236]3 years ago

7 0

If I remember correctly, and I might be wrong, I think that the answer is D. I hope this helps you. :)

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Point B ∈ AC. Find the length of segments AB and BC if AC = 20 cm and: AB:BC = 3:7

madam [21]

Answer:

Length of AB = 6 cm

Length of the segment BC = 14 cm

Step-by-step explanation:

Here, B is a point on a segment AC.

AB : BC = 3:7

Length of the segment AC = 20 cm

Now, let the common ratio between the segment is x.

So, the length of AB = 3 x , and Length of BC = 7 x

Now, AB + BC = AC

⇒ 3x + 7x = 20

or, 10 x = 20

or, x = 2

Hence, the length of AB = 3 x = 3 x 2 = 6 cm

and the length of the segment BC = 7x = 7 x 2 = 14 cm

7 0

3 years ago

Read 2 more answers

Use ABC to find the value of sin A. a. 12/37 b. 37/12 c. 35/37 d. 12/35

FromTheMoon [43]

Answer:

a. 12/37

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you ...

Sin = Opposite/Hypotenuse

sin(A) = BC/AB = 12/37

4 0

3 years ago

2 096 to the nearest 1000

k0ka [10]

The answer to this is 2000...
hope this helps and HAGD!

6 0

2 years ago

Read 2 more answers

Suppose that an airline overbooks seats on their flights. In particular, it sells 300 tickets for a flight when there are only 2

vladimir1956 [14]

Using the normal approximation to the binomial, it is found that there is a 0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

15% do not show up, so 100 - 15 = 85% show up, which means that $p = 0.85$ .
300 tickets are sold, hence $n = 300$ .

The mean and the standard deviation are given by:

$\mu = np = 300(0.85) = 255$

$\sigma = \sqrt{np(1-p)} = \sqrt{300(0.85)(0.15)} = 6.185$

The probability that we will have enough seats for everyone who shows up is the probability of at most 270 people showing up, which, using continuity correction, is $P(X \leq 270 + 0.5) = P(X \leq 270.5)$ , which is the p-value of Z when X = 270.5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{270.5 - 255}{6.185}$

$Z = 2.51$

$Z = 2.51$ has a p-value of 0.994.

0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

A similar problem is given at brainly.com/question/24261244

8 0

3 years ago

In what follows use any of the following tests: Regression, multiple regression, one-sided t-test, or two-sided t-test. All conc

Yakvenalex [24]

Answer:

Two sided t-test ( d )
0.245782 ( c )
Since P-value is too large we cannot conclude that the students’ weight are different for these two schools. ( c )
The test is inconclusive; thus we cannot claim that the average weights are different. ( b )

Step-by-step explanation:

1) Test performed is a Two sided test and this because we are trying to determine the mean difference between two groups irrespective of their direction

2) Determine the P-value ( we will use a data-data analysis approach on excel data sheet while assuming Unequal variances )

yes No

Mean 94.47059 89.76471

Variance 173.2647 95.19118

Observations 17 17

df 30

t Stat 1.184211

P(T<=t) one-tail 0.122814

t Critical one-tail 1.697261

P(T<=t) two-tail 0.245782

Hence The p-value = 0.245782

3) Since P-value is too large we cannot conclude that the students’ weight are different for these two schools.

4) The test is inconclusive; thus we cannot claim that the average weights are different.

8 0

2 years ago