What is 3.1666666667 as a fraction

It is 4.7 km from Lighthouse A to Port B. The bearing of the port from the lighthouse is N73°E. A ship has sailed due west from

Gennadij [26K]

Answer:

correct option is C. 3.7 km

Step-by-step explanation:

given data

A to Port B = 4.7 km

lighthouse = N73°E

lighthouse = N31°E

solution

we get here first $\angle$ B and

here

$\angle$ A = 90 - 73 = 17°

$\angle$ B = 73 - 31 = 42°

and

sum of all angle 180° so

$\angle$ A +

17° + 42° + $\angle$ C = 180°

solve it we get

$\angle$ C = 121°

Now we use here sin law that is

$\frac{b}{sinB} = \frac{c}{sinC}$ ........................2

put here value and we get

$\frac{b}{sin42} = \frac{4.7}{sin121}$

solve it we get

b = 3,7 km

so correct option is C. 3.7 km

7 0

1 year ago

Find b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC

Kamila [148]

<span>The side b is opposite to the angle B, applying the law of the sines, we have:

</span> $\frac{a}{sinA} = \frac{b}{sinB}$
$\frac{20}{sin30^0} = \frac{b}{sin45^0}$
$\frac{20}{ \frac{1}{2} } = \frac{b}{ \frac{ \sqrt{2} }{2} }$
$20* \frac{ \sqrt{2} }{2} = b* \frac{1}{2}$
$\frac{20 \sqrt{2} }{2} = \frac{b}{2}$
$2*b =2*20 \sqrt{2}$
$2b = 40 \sqrt{2}$
$b = \frac{40 \sqrt{2} }{2}$
$\boxed{b = 20 \sqrt{2} }$

6 0

1 year ago

Read 2 more answers

Max travels to see his brother's family by car. He drives 216 miles in 4 hours.

LUCKY_DIMON [66]

Answer:

54 mph

Step-by-step explanation:

3 0

2 years ago

The figure shows the design of a greenhouse with a rectangular floor. The front and back sides of the greenhouse are semicircles

Ksju [112]

9514 1404 393

Answer:

96 cubic feet

Step-by-step explanation:

A volume that is 16 ft by 18 ft by 1/3 ft will be ...

V = LWH

V = (16 ft)(18 ft)(1/3 ft) = 96 ft³

96 cubic feet of concrete are needed.

7 0

2 years ago

A study was performed on green sea turtles inhabiting a certain area. Time-depth recorders were deployed on 6 of the 76 capture

polet [3.4K]

Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean shell length = 51.3 cm

s = sample standard deviation = 6.6 cm

n = sample of turtles = 6

$\mu$ = true mean shell length

Now, the 99% confidence interval for $\mu$ = $\bar X \pm t_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$

Here, $\alpha$ = 1% so $(\frac{\alpha}{2})$ = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> $\mu$ = $51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }$

= [51.3 - 10.864 , 51.3 + 10.864]

= [40.44 cm, 62.16 cm]

The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

3 0

1 year ago