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3 years ago

5

If sin theta = 4/5 then sec theta is?

1 answer:

Softa [21]3 years ago

6 0

Answer: 53.13
Explanation of answer: since theta is the unknown, then you must use the inverse of sin. (Which is sin with -1 subscript) I used a calculator and used inverse of sin and put 4/5 in parenthesis

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Bonsoir ce que vous pouvez m aidez <br> je cherche des fraction égale a 4sur16

musickatia [10]

Answer:

1 sur 4 est équivalent.

Step-by-step explanation:

J'espère que cela aide !! Puis-je avoir la couronne?

4/16 et être réduit à 1/4!

6 0

3 years ago

Solve:-<br><br> 7^5<br><br> Thanks!!!!!!!!!!!!!

solong [7]

These are exponents. To solve an exponent you need to multiply the base it self how many times its says in the power.

a²

a = base
<span>2 = power
</span>
7^5 = 7 × 7 × 7 × 7 × 7 = <span>16807
7^5 = </span><span>16807</span>

3 0

3 years ago

Read 2 more answers

3 (x+2)= 6 (x-1) +3

zzz [600]

Answer:

x = 3

Step-by-step explanation:

3(x + 2) = 6(x - 1) + 3

Divide both sides by 3.

x + 2 = 2(x - 1) + 1

Distribute the 2 on the right side.

x + 2 = 2x - 2 + 1

Combine like terms on the right side.

x + 2 = 2x - 1

Add 1 to both sides. Subtract x from both sides.

3 = x

x = 3

3 0

3 years ago

Write a real world problem for: f + 5 = 9

Triss [41]

9-5=4 Therefore f=4
Hope this helps

4 0

3 years ago

Read 2 more answers

The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand

Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

$P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})$

$P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})$

$P(X > 1560) = P(Z > \dfrac{110}{220})$

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean $\mu_x$ = np = 52 × 0.3085 = 16.042

The standard deviation = $\sqrt {n \time p (1-p)}$

The standard deviation = $\sqrt {52 \times 0.3085 (1-0.3085)}$

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

$Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})$

$Pr ( Y > 20) = P(Z >1 .338)$

From z tables

P(Y > 20) $\simeq$ 0.0903

7 0

3 years ago