All categories

2 years ago

Find the angle whose cosine is .4203

Mathematics

1 answer:

ludmilkaskok [199]2 years ago

5 0

The cosine of 1.137020412 degrees is 0.4204.

You might be interested in

CAn anyone help me with this question plz

mario62 [17]

Answer:

3000ft²

Step-by-step explanation:

2cm =5 ft

24 cm = 60 ft

20 cm = 50 ft

Area = 60 *50

= 3000

3 0

2 years ago

A garden is in the shape of a semicircle with a radius of 25 feet. Fencing is placed around this garden. How much fencing, in fe

Len [333]

Answer:

S.A. = 981.7 ft^2

Step-by-step explanation:

radius = 25 ft

S.A. = 1/2πr2

S.A. = 1/2π(25)^2

S.A. = 981.7

4 0

3 years ago

Read 2 more answers

Write an equation in standard form of the line that contains the point (-1,2) amd is perpendicular to y = 3x - 1

Strike441 [17]

Since it is perpendicular to y=3x-1, the slope must be a negative reciprocal.
there the slope of the perpendicular line is $-\frac{1}{3} x$

So now we have,
$y=- \frac{1}{3} x+b$

To find b, we use the point that the line passes through and substitute it to the y and x values.

$2=- \frac{1}{3} (-1)+b$

which would give us b=5/3

Therefore, the equation of the line is
$y=- \frac{1}{3} x+ \frac{5}{3}$

7 0

2 years ago

a paddle boat can move at a speed of 6km/h in still water. the boat is paddeled 16km downstream in a river in the same time it t

kaheart [24]

$v-the\ speed\ of\ the\ river\\\\16:(6+v)=8:(6-v)\\\\\frac{16}{6+v}=\frac{8}{6-v}\\-------------\\multiply\ on\ the\ cross\\-------------\\8(6+v)=16(6-v)\\\\48+8v=96-16v\\\\8v+16v=96-48\\\\24v=48\ \ \ /:24\\\\v=2\ (km/h)\leftarrow Answer$

5 0

2 years ago

The number of text messages sent daily by a student is a poisson random variable with parameter λ=5 .in a class with 20 independ

Rudik [331]

This problem is a combination of the Poisson distribution and binomial distribution.

First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961

For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
=6.18101*10^(-5) or approximately
=0.00006181

7 0

3 years ago