All categories

2 years ago

Find the area of the triangle. Give the answer to the nearest tenth. The drawing may not be to scale.

Mathematics

1 answer:

LenaWriter [7]2 years ago

3 0

Answer:

10.5 m²

Step-by-step explanation:

Given:

∆ with 2 side lengths, 4.7m, 6.1m, and an included angle, 47°.

Required:

Area of the triangle

SOLUTION:

Area of such triangle that has 2 known sides and an included angle between them is given as ½*a*b*sin(θ).

Where,

a = 4.7 m

b = 6.1 m

θ = 47°

Area = ½*4.7*6.1*sin(47)

Area = 10.5 m² (to the nearest tenth)

You might be interested in

Solve the equation -10 (g - 6) = -65

Elanso [62]

Step-by-step explanation:

$- 10(g - 6) = - 65 \\ - 10g + 60 = - 65 \\ - 10g = - 125 \\ 10g = 125 \\ g = \frac{25}{2}$

5 0

2 years ago

XY=7a,YZ=5a,XZ=6a+24

leonid [27]

7a + 5a = 6a + 24
6a = 24 
a = 4 
XZ = 6(4) + 24 
XZ = 48

5 0

2 years ago

The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able

konstantin123 [22]

Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by $t_1=$ 2.1 minutes.

Then they are able to reduce that time by $t_2=$ 10

If their final time is 3.96 minutes, then

$x-t_1-t_2=3.6\\\Rightarrow\ x=3.6+t_1+t_2\\\Rightarrow\ x=3.6+2.1+10\\\Rightarrow\ x= 15.7$

Hence, their beginning time was 15.7 minutes.

7 0

2 years ago

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.

In-s [12.5K]

Answer:

The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=15,\sigma=4)$