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dybincka [34]
3 years ago
11

Help please thanks don’t know how to do this

Mathematics
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

a = 11.71 ; b = 15.56

Step-by-step explanation:

For this problem, we need two things.  The law of sines, and the sum of the interior angles of a triangle.

The law of sines is simply:

sin(A)/a = sin(B)/b = sin(C)/c

And the sum of interior angles of a triangle is 180.

45 + 110 + <C = 180

<C = 25

We can find the sides by simply applying the law of sines.

length b

7/sin(25) = b/sin(110)

b = 7sin(110)/sin(25)

b = 15.56

length a

7/sin(25) = a/sin(45)

a = 7sin(45)/sin(25)

a = 11.71

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Consider the following points.
horrorfan [7]

Answer:

(a) P(x)=0.4x^{3} +2x^{2} -1.4x

(b)P(x)=x^{4} -x^{3} +5x^{2} -7x-14

Step-by-step explanation:

Let´s use Divided Differences Method of Polynomial Interpolation given by this iteration:

f[x_k,x_k_+_1,...,x_k_+_i]=\frac{f[x_k_+_1,x_k_+_2,...,x_k_+_i]-f[x_k,x_k_+_1,...,x_k_+i_-_1]}{x_k_+_i-x_k}

k∈[0,n-i]

Thus the Newton polynomial can be written as:

P_n_-_1(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_o,x_1,x_2](x-x_0)(x-x_1)+...+f[x_n,x_n_-_1,...,x_1](x-x_n)(x-x_n_-_1)...(x-x_1)

(a) I attached you the procedure in the first table, using it we have:

P(x)=3+(-3)(x+1)+(2)(x+1)(x)+0.4(x+1)(x)(x-1)

Operate P(x) using the distributive property:

P(x)=0.4x^{3} +2x^{2} -1.4x

(b) I attached you the procedure in the second table, using it we have:

P(x)=44+(-44)(x+2)+(15)(x+2)(x+1)+(-3)(x+2)(x+1)(x)+(1)(x+2)(x+1)(x)(x-1)

Operate P(x) using the distributive property:

P(x)=x^{4} -x^{3} +5x^{2} -7x-14

6 0
3 years ago
A simple random sample from a population with a normal distribution of 102 body temperatures has x overbarequals98.40degrees Upp
Oliga [24]

Answer:

It is 80% statistically safe to conclude that the population standard deviation is less than 1.8°F

Step-by-step explanation:

The given information are;

The sample size, n = 102

The sample mean = 98.4°F

The sample standard deviation = 0.66°F

\sqrt{\dfrac{\left (n-1  \right )s^{2}}{\chi _{\alpha /2}^{}}}< \sigma < \sqrt{\dfrac{\left (n-1  \right )s^{2}}{\chi _{1-\alpha /2}^{}}}

α = 0.2, ∴ α/2 = 0.1

\chi _{1-\alpha /2} = \chi _{0.9, 101} = 83.267

\chi _{\alpha /2} = \chi _{0.1, 101} = 119.589,

Which gives;

\sqrt{\dfrac{\left (102-1  \right )0.66^{2}}{119.589}^{}}}< \sigma < \sqrt{\dfrac{\left (102-1  \right )0.66^{2}}{83.267}^{}}}

0.607 < σ <0.727

Therefore, it is 80% statistically safe to conclude that the population standard deviation is less than 1.8°F.

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