Square ABCD and isosceles triangle EFG have the same perimeter. Work out the length of FG.

Suppose abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume th

HACTEHA [7]

Given that:
mean,μ=35.6 min
std deviation,σ=10.3 min
we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.
using z-table, the z-score that will give us 0.2296 is:-1.99
therefore:
z-score is given by:
(x-μ)/σ
hence:
-1.99=(x-35.6)/10.3
-20.497=x-35.6
x=35.6-20.497
x=15.103

7 0

2 years ago

Read 2 more answers

Mary Hernandez had a policy with a $250 deductible which paid 80% of her covered charges less deductible. She had medical expens

Nataly_w [17]

A. the insurance company's payment = $14,392
$18240 - $250 = 17990 x 80% = $14,392

b. <span>the 20% copayment
17990 * 20% = $3598
</span>
<span>c. Mary's total cost
$250 + $3598 = $3848
</span>

6 0

2 years ago

Read 2 more answers

Is a diameter of circle O. If m∠ACB = 30°, what is m∠DBA? A. 30° B. 50° C. 60° D. 70° E. 90°

lesya692 [45]

Every triangle equals 180 degrees, and we're trying to find the angle of DBA. So, we can identify DAB as a 90 degree angle to start. Then we can identify that ACB = ADB, meaning ADB is 30 degrees. With that information, we have this equation:

DBA = 180 - 90 - 30

DBA = 60

6 0

2 years ago

James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo

Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is $y = 2 ^n$

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒ n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒ for n = 1 , y = 2 ..... (1)

Similarly, if the paper if folded TWICE ⇒ n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒ for n = 2 , y = 4 ..... (2)

⇒ $y = 2^2 = 2^n$

Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7

So, $y = 2^ n = 2^7 = 128$

⇒ There are total 128 equal parts.

Lastly, if the paper is folded ELEVEN times ⇒ n = 11

So, $y = 2^ n = 2^{11} = 2048$

⇒ There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is $y = 2 ^n$

8 0

2 years ago

Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a fl

BigorU [14]

You have two 30-60-90 triangles, ADC and BDC.

The ratio of the lengths of the sides of a 30-60-90 triangle is

short leg : long leg : hypotenuse
1 : sqrt(3) : 2

Using triangle ADC, we can find length AC.
Using triangle BDC, we can find length BC.
Then AB = AC - BC

First, we find length AC.
Look at triangle ACD.
DC is the short leg opposite the 30-deg angle.
DC = 10sqrt(3)
AC = sqrt(3) * 10sqrt(3) = 3 * 10 = 30

Now, we find length BC.
Look at triangle BCD.
For triangle BCD, the long leg is DC and the short leg is BC.
BC = 10sqrt(3)/sqrt(3) = 10

AB = AC - BC = 30 - 10 = 20

6 0

2 years ago

Read 2 more answers