All categories

3 years ago

62 62 65 65 115 115 118 118 I have no idea how to solve this, will give brainliest

Mathematics

1 answer:

Kobotan [32]3 years ago

8 0

Answer:

1.ABC = 45°,DBE = 45°. ABD = 135°,CBE = 135°

2.ABC = 62°, DBE = 62°, ABD =118°,CBE = 118°

Step-by-step explanation:

1.angle ABC = angle DBE because they are vertical. You can write the equation 3x+38 = 5x+20 and find both angles. When you solve, x=9. You substitute it in. ABD is equal to 180-45° because it is supplementary with ABC. ABD = CBE b/c vertical angles.

2.You do the same thing. 4x+2=5x-13,x=15. You substitute in 15 for x and get 62°. They are the equal angles. For ABD you do 180 -62 since they are supplementary angles, which gets you 118 which applies for both because they are vertical angles.

I want the brainliest.

You might be interested in

Shirt = $14.99 pants = $27.00 sweater = $36.99 About how much $ in all?

MissTica

Answer: $78.98, about $79

Step-by-step explanation:

14.99+27+26.99=78.98

7 0

3 years ago

Read 2 more answers

Seventy weather balloons were sent up. Twenty-eight were found. What percent wer ound? 17

In-s [12.5K]

Step-by-step explanation:

let the percent of found balloons be x

given,

total number of balloons= 70

number of found balloons= 28

according to the question,

x% of 70= 28

x/100 × 70 = 28

x=40

hence 40% of the balloons were found

5 0

3 years ago

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$

$P(3 \leq x \leq 4) = 23.7\%$

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85$

$= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85$

$=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15$

Calculation the value from standard normal z table, we have,

$P( z \leq -1.036) = 0.15$

$\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65$

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0

3 years ago

Karel sells goods to customers in exchange for a $100,000 noninterest-bearing note due in 2 years. The interest rate on this typ

Zielflug [23.3K]

Answer:

$85,734

Step-by-step explanation:

We have been given that Karel sells goods to customers in exchange for a $100,000 non-interest-bearing note due in 2 years. The interest rate on this type of loan is 8%. We are asked to find the present value of the note.

$PV=FV\frac{1}{(1+r)^n}$ , where,