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2 years ago

324 x 46 first partial product

Mathematics

1 answer:

Morgarella [4.7K]2 years ago

7 0

Answer

1944(324x6)+12960(324x40) or 14904

Step-by-step explanation:

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Textbook: Please write as an expression involving functions of Θ alone.

Setler [38]

Step-by-step explanation:

tan(Θ + 30°)

Use angle sum formula:

tan(A + B) = (tan A + tan B) / (1 − tan A tan B)

So we can write this as:

(tan Θ + tan 30°) / (1 − tan Θ tan 30°)

tan 30° = 1/√3, so:

(tan Θ + 1/√3) / (1 − 1/√3 tan Θ)

Multiply top and bottom by √3:

(√3 tan Θ + 1) / (√3 − tan Θ)

6 0

3 years ago

Can some one help me plz

tigry1 [53]

B but the last one in the b row

5 0

2 years ago

Read 2 more answers

Y=2x^2 - 10x + 13<br> Find the value.

pishuonlain [190]

Image of graph The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

6 0

2 years ago

An arithmetic sequence is defined by the recursive formula t1 = 11, tn = tn - 1 - 13, where n ∈N and n > 1. Which of these is

galben [10]

coolio

$t_1=11$

$t_n=t_{n-1}-13$

so each term is ound by subtracting 13 from the previous term

an aritmetic sequence can be written as

$t_n=t_1+d(n-1)$ were

$t_n$ is the nth term

$t_1$ is the first term

d is common difference, which can also be found by doing $t_n-t_{n-1}=d$

n=wich term

we know that $t_1=11$ and we can find d

$t_n=t_{n-1}-13$ , $t_n-t_{n-1}=-13=d$

so te general term is $t_n=11-13(n-1)$ which can also be expanded and written as $t_n=-13n+24$

6 0

3 years ago

birth weights in norway are normally distributed with a mean of 3570 g and a standard deviation of 500g. if the hospital officia

Svetradugi [14.3K]

Answer:

2630 g

Step-by-step explanation:

From the given information:

Given that:

mean (μ) = 3750 g

Standard deviation (σ) = 500

Suppose the hospital officials demand special treatment with a percentage of lightest 3% (0.03) for newborn babies;

Then, the weight of birth that differentiates the babies that needed special treatment from those that do not can be computed as follows;

P(Z < z₁) = 0.03

Using the Excel Formula; =NORMSINV(0.03) = -1.88

z₁ = - 1.88

Using the test statistics z₁ formula:

$z_1 = \dfrac{X-\mu}{\sigma}$

$-1.88 = \dfrac{X-3570}{500}$

By cross multiply, we have:

-1.88 × 500 = X - 3570

-940 = X - 3570

-X = -3570 + 940

-X = -2630

X = 2630 g

Hence, 2630 g is the required weight of birth that differentiates the babies that needed special treatment from those that do not

6 0

2 years ago