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4 years ago

Trigonometry heeellllppp plzzzz

Mathematics

1 answer:

algol [13]4 years ago

3 0

Answer:

x ≈ 3.4 cm

Step-by-step explanation:

Using the sine ratio in the right triangle

sin57° = $\frac{opposite}{hypotenuse}$ = $\frac{x}{4}$ ( multiply both sides by 4 )

4 × sin57° = x , thus

x ≈ 3.4 cm ( to 1 dec. place )

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If eight cars are entered in a race and three finishing places are considered, how many finishing orders can they finish? Assume

Sladkaya [172]

Answer:

a) 336

b)6

c)6

Step-by-step explanation:

a) In the 1st place fit 8 cars, in the 2nd only fit 7 cars, because one is already in the first place and 3th fit 6 cars, so all the possibilities are 8*7*6.

b) in the 1st place fit 3 cars, 2nd 2 cars and 3th 1 car, so all the possibilities are 3*2*1

c) every shirt can be matched with 1 pair of pants, there are 2 shirts and 3 pants, so all the the possibilities are 6+6.

3 0

4 years ago

How and what???? I need the answer to #2

allochka39001 [22]

0,7 1,8 5,1 6,7 .... this is a function as well!!

6 0

3 years ago

Which graph represents this system? 2x-5y =-5 y={x+1

Yuri [45]

Answer:

Step-by-step explanation:

4 0

3 years ago

Consider the following sample set of scores. Assume these scores are from a discrete distribution. 21 29 32 38 38 45 50 64 72 10

ASHA 777 [7]

Answer:

Original data: 21 29 32 38 38 45 50 64 72 100

$\bar X = \frac{21+29+32+38+38+45+50+64+72+100}{10}=48.9$

Mode =38.

$Median = \frac{38+45}{2}=41.5$

Change 1: 2 29 32 38 38 45 50 64 72 100

$\bar X = \frac{2+29+32+38+38+45+50+64+72+100}{10}=47$

Mode =38.

$Median = \frac{38+45}{2}=41.5$

Change 2: 29 32 38 38 45 50 64 72 100

$\bar X = \frac{29+32+38+38+45+50+64+72+100}{9}=52$

Mode =38.

$Median = 45$

Step-by-step explanation:

Original data: 21 29 32 38 38 45 50 64 72 100

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{10} X_i}{10}$

And if we replace we got:

$\bar X = \frac{21+29+32+38+38+45+50+64+72+100}{10}=48.9$

The mode is the most repeated value in the sample and on this case is Mode =38.

Since we have an even number of points the median is calculated as the average between the observations 5 and 6 from the dataset ordered.

$Median = \frac{38+45}{2}=41.5$

Change 1: 2 29 32 38 38 45 50 64 72 100

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{10} X_i}{10}$

And if we replace we got:

$\bar X = \frac{2+29+32+38+38+45+50+64+72+100}{10}=47$

The mode is the most repeated value in the sample and on this case is Mode =38.

Since we have an even number of points the median is calculated as the average between the observations 5 and 6 from the dataset ordered.

$Median = \frac{38+45}{2}=41.5$

Change 2: 29 32 38 38 45 50 64 72 100

Now the sample size is 9 instead of 10

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{9} X_i}{9}$

And if we replace we got:

$\bar X = \frac{29+32+38+38+45+50+64+72+100}{9}=52$

The mode is the most repeated value in the sample and on this case is Mode =38.

Since we have odd number of points the median is calculated from the 5 position of the dataset ordered.

$Median = 45$

7 0

4 years ago

I NEED HELP ASAP

SIZIF [17.4K]

Answer:

It is not reasonable that the state education department claims the percentage for the entire state is 73%.

Step-by-step explanation:

We are given that 191 of the 288 high school students surveyed at a local school said they went outside more during school hours as elementary school students than they do now as high school students.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of high school students who went outside more during school hours as elementary school students than they do now as high school students = $\frac{191}{288}$ = 0.66

n = sample of high school students = 288

p = population percentage for the entire state

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

The margin of error is given by;

M.E. = $2 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

= $2 \times \sqrt{\frac{0.66(1-0.66)}{288} }$

M.E. = 0.056 or 5.6%

So, the confidence interval so formed = $\hat p \pm \text{Margin of error}$

= [ $0.66 - 0.056 , 0.66 + 0.056$ ]

= [0.604, 0.716]

Since the above interval does not include 0.73 or the population proportion of 73% falls outside the above interval. So, it is not reasonable that the state education department claims the percentage for the entire state is 73%.

4 0

3 years ago

Trigonometry heeellllppp plzzzz​

Trigonometry heeellllppp plzzzz