All categories

3 years ago

If t=38.5 and s=31.4 find S. Round to the nearest tenth

Mathematics

2 answers:

zubka84 [21]3 years ago

8 0

<u>Answer:</u>

The correct answer option is C. S = 54.6°.

<u>Step-by-step explanation:</u>

We are given a right angled triangle with two known sides, $s$ and $t$ .

We are to find the value of the angle $S$ .

For that, we will use sine.

$sin S = \frac { s } { t }$

$sin S = \frac { 3 1 . 4 } { 3 8 . 5 }$

$S = sin'0.815$

S = 54.6°

Dima020 [189]3 years ago

6 0

Answer: option c.

Step-by-step explanation:

You need to remember the identity:

$sin\alpha=\frac{opposite}{hypotenuse}$

The inverse of the sine function is arcsine. You need to use this to calculate the angle "S":

$\alpha =arcsin(\frac{opposite}{hypotenuse})$

You know that you need to find the measure of "S" and $t=38.5$ (which is the hypotenuse) and $s=31.4$ (which is the opposite side), then you can substitute values into $\alpha =arcsin(\frac{opposite}{hypotenuse})$

Then, you get:

$S=arcsin(\frac{31.4}{38.5})\\\\S=54.6\°$

You might be interested in

Find the probability of the following events , when a dice is thrown once:

Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

7 0

3 years ago

Read 2 more answers

Please help!! Will mark brainliest

Viefleur [7K]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The U.S. Bureau of Labor Statistics reports that of persons who usually work full-time, the average number of hours worked per w

alukav5142 [94]

Answer:

The standard deviation of number of hours worked per week for these workers is 3.91.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

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

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem we have that:

The average number of hours worked per week is 43.4, so $\mu = 43.4$ .

Suppose 12% of these workers work more than 48 hours. Based on this percentage, what is the standard deviation of number of hours worked per week for these workers.

This means that the Z score of $X = 48$ has a pvalue of 0.88. This is Z between 1.17 and 1.18. So we use $Z = 1.175$ .