1+1 then divide by 8

The weight of oranges growing in an orchard is normally distributed with a mean weight of 6.5 oz. and a standard deviation of 1.

sergey [27]

Add 3 standard deviations above and below the mean to get the range in which 99.7% of the data in a normal distribution will fall
6.5 + 4.5 = 11
6.5 - 4.5 = 2
So 2 to 11 ounces would be the interval

6 0

2 years ago

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

or

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

7 0

3 years ago

Jasmine loves to babysit. On Saturday, she babysat from 3:29 p.m. to 4:52 p.m. On Sunday, she babysat from 1:37 p.m. to 3:16 p.m

patriot [66]

In Sunday she babysits 1 hour and 23 minutes,in Saturday she babysit 1 hour and 39 minutes.In total it’s 3 hours and 12 minutes.

4 0

3 years ago

Solve with explanation -4x+5> - 5

Bezzdna [24]

Answer:

$x < 2.5$

Step-by-step explanation:

$- 4x + 5 > - 5 \\ - 4x > - 5 - 5 \\ - 4x > - 10 \\ \frac{ - 4x}{ - 4} > \frac{ - 10}{ - 4} \\ x < \frac{5}{2} \\ x < 2.5$

hope this helps

brainliest appreciated

good luck! have a nice day!

7 0

3 years ago

Which expression is equivalent to one over five m − 20? (4 points)

exis [7]

Answer:

The equivalent to one over five m − 20 is one over five (m − 100) ⇒ B

Step-by-step explanation:

Let us solve the question

∵ One over five means $\frac{1}{5}$

∴ One over five m - 20 = $\frac{1}{5}$ m - 20

→ By using the distributive property, take one over five as a common factor

from both terms

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$

→ Simplify the bracket

∵ $\frac{1}{5}$ m ÷ $\frac{1}{5}$ = $\frac{1}{5}$ m × 5 = m

∵ 20 ÷ $\frac{1}{5}$ = 20 × 5 = 100

∴ $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$ = $\frac{1}{5}$ (m - 100)

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ (m - 100)

The equivalent to one over five m − 20 is one over five (m − 100)

4 0

4 years ago