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2 years ago

Math I need help I with it if you can help me out

Mathematics

1 answer:

e-lub [12.9K]2 years ago

4 0

Answer: 49.44807

Round that however you need to. Unfortunately the rounding instructions on the screenshot are cut off, so I cannot see what that part says.

=======================================================

Work Shown:

sin(angle) = opposite/hypotenuse

sin(22) = h/132

h = 132*sin(22)

h = 49.44807

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A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0

4 years ago

They have a rectangle ABCD as shown below, where AB = 24cm BC = 10cm and AE = 13cm

alex41 [277]

Answer:

A. Area of ABCD = 240 $cm^{2}$

B. 60 cm

C. 36 cm

D. 50 cm

Step-by-step explanation:

Given: AB = 24cm BC = 10cm and AE = 13cm.

A. Since a rectangle is a 2 dimensional figure, it has no volume but area.

So that,

the area of the rectangle ABCD = length x width

= 24 x 10

= 240 $cm^{2}$

B. To calculate the circumference of the BCD triangle, apply the Pythagoras theorem to determine BD.

$BD^{2}$ = $BC^{2}$ + $DC^{2}$

= $10^{2}$ + $24^{2}$

= 676

BD = $\sqrt{676}$

= 26

BD = 26 cm

so that,

the circumference of BCD = 10 + 24 + 26

= 60 cm

C. To calculate the circumference of the BEC triangle,

AC = 26 cm, AE = 13 cm

CE = 26 - 13

= 13 cm

CE = 13 cm

The circumference of the BEC triangle = 13 + 13 + 10

= 36 cm

D. The circumference of the DEC triangle = 13 + 13 + 24

= 50 cm

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