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2 years ago

What is the length of the diagonal from B to D of this rectangle? Round your answ

Mathematics

1 answer:

pishuonlain [190]2 years ago

5 0

Answer: 19.7

Step-by-step explanation:

By the Pythagorean theorem,

$(BD)^{2}=8^{2}+18^{2}\\\\(BD)^{2}=388\\\\BD=\sqrt{388} \approx \boxed{19.7}$

You might be interested in

If you deposit $6000 into an account paying 8% annual interest compounded monthly, how much money will be in the account after 7

Leviafan [203]

Answer:

$10278

Step-by-step explanation:

Given data

P= $6000

R= 8%

T= 7 years

The compound interest formula is

A=P(1+r)^t

substitute

A=6000(1+0.08)^7

A=6000(1.08)^7

A=6000*1.713

A=$10278

6 0

2 years ago

If MB = 6x-9, and AM = 27, find x.

telo118 [61]

Answer:

x=11

Step-by-step explanation:

M is the midpoint this means that we can set AM and BM equal to eachother, once this is done we can solve for x

9x-6=6x+27 (set equal to eachother)

9x=6x+33 (add 6)

3x=33 (subtract 6x)

x=11 (divide by 3)

7 0

3 years ago

What is 882992+2899992237×2222÷29727

maw [93]

Answer:

217648314.791

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Helpppppppppppppppppp

mariarad [96]

Answer:

$2400

$12,400

Step-by-step explanation:

Simple interest equation:

PV(1+it)

we have

10000(1+.06*4)

10000(1.24)

12400

Subtract the PV to get the interest earned

12400-10000= 2400

We just found that the total would be 12,400 and so part B is already solved

4 0

3 years ago

Assume a standard deviation of LaTeX: \sigma = 0.75σ = 0.75. You plan to take a random sample of 110 households, what is the pro

olchik [2.2K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, $\mu$ = 2.58 and Standard deviation, $\sigma$ = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar $\leq$ 2.50)

P(Xbar < 2.66) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z < -1.68) = 1 - P(Z $\leq$ 1.68)

= 1 - 0.95352 = 0.04648

P(Xbar $\leq$ 2.50) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z $\leq$ -3.92) = 1 - P(Z < 3.92)

= 1 - 0.99996 = 0.00004

Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046

3 0

3 years ago