Please look at a the attached image below, it’s urgent

Kiara's birthday celebration started at 2:00 pm. She spent 46 minutes at the trampoline park, Eating cake and opening presents t

Phantasy [73]

Answer:

Im not 100% sure but I got 1.66667 so therefore I would say about 4:00 maybe? Didn't really understand lmk.

Step-by-step explanation: I added 46, 29, and 25 together which gave me 100 and i put that into minutes which gave me 1.66667 which rounds to 1.70 and so if 60 seconds is 1 hour than 2:00 plus 2 hours is 4. That's what I got out of it.

5 0

3 years ago

Prove : (sec θ - tan θ )^2 = 1 - sin θ /1+sin θ

labwork [276]

$(sec x - tan x)^2 \\ \\ = sec^2x - 2 sec x tan x + tan^2 x \\ \\ =(1+tan^2 x) - 2 sec x tan x +tan^2 x \\ \\ =1 - 2 sec x tan x + 2 tan^2 x \\ \\ = 1 - 2tan x(sec x - tan x) \\ \\ =1 - \frac{2 sin x}{cos x} (\frac{1-sin x}{cos x}) \\ \\ = 1 - \frac{2 sin x (1-sin x)}{cos^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{1-sin^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{(1-sin x)(1+sin x)} \\ \\ =1-\frac{2 sin x}{1+sin x}$

3 0

3 years ago

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

5 0

3 years ago

Which interval notation accurately represents the domain?

Sauron [17]

Answer:

the 1st one

next

the 3rd one

Step-by-step explanation: edg 2020

7 0

3 years ago

What's is the answer

Anna71 [15]

The answer is #2 you’re welcome

5 0

2 years ago