How is 0.009 and 0.9 related

In aΔABC, if ∠B = 60° and the ratio of two sides is a : c = 2 : √3 + 1, then ∠A= ____.

UkoKoshka [18]

Answer:

∠A= 45°....

Step-by-step explanation:

To find ∠A first we have to find the length of side B by cosine law:

b^2 = a^2 + c^2 – 2 a c cos B

b^2=(2)^2 +(√3 + 1)^2 - 2(2)(√3 + 1) cos 60

b^2 = 6

Taking square root at both sides:

√b^2 = √6

b= 2.45

Now we can calculate ∠A by sine law:

b / sin B = a / sin A

2.45 / sin 60 = 2 / sin A

sin A= 2* √3/2 /2.45

sin A = 2√3/2 * 1/2.45

sinA = 2√3/ 4.9

sin A = 0.7069

sin A = 0.707

A=45°

Thus ∠A= 45°....

4 0

3 years ago

I could use some help with this problem.

Alla [95]

Answer:

A

Step-by-step explanation:

Plug in -6 to the equation

-5 - 3(-6) > 10

Simplify for -5 + 18 = 13

13 > 10

4 0

3 years ago

Read 2 more answers

The sum of 17 and a number x is expand to 20 (Write an equation for the decription)

Rudiy27

So the sum of 17 and a number x means to add them on one side of the equals sign. The other side would be your 20.

17 + x = 20

3 0

3 years ago

Please please pleaseeeeeeeeeeeeeeeeeee I beg of your assistance

hoa [83]

Answer:

80/160

Step-by-step explanation:

so basically...

50 is half of 100

so then you look at it and your like welp if its half then ur going to have to do half for 80

so 80 x 2= 160

and for A can you please show the options or at least tell me

7 0

3 years ago

Read 2 more answers

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .