75,300,207 in word form

Find the total mass of the triangular region shown below. All lengths are in cm, and the density of the region is sigma(x) = 1 +

anastassius [24]

Answer:

3 grams

Step-by-step explanation:

We are going to take the mass of a bunch of little strips below the triangle "roof." To do this, we must figure out what formula for the mass we'll use, in this case, we'll use:

Mass of strip = denisty * area = (1+x)*y*deltax grams

now, because the "roof" of the triangle contains two different integrals (it completely changes direction), we will use TWO integrals!

**pretend ∈ is the sum symbol

Mass of left part = lim x->0 ∈ (1+x)*y*deltax = inegral -1 to 0 of (1+x)*3*(x+1) = 3 * integral -1 to 0 of (x^2 + 2x + 1) = 3 * 1/3 = 1

Mass of left part = lim x->0 ∈ (1+x)*y*deltax = inegral 0 to 1 of (1+x)*3*(-x+1) = 3 * integral 0 to 1 of (-x^2 + 1) = 3 * 2/3 = 2

Total mass = mass left + mass right = 1 + 2 = 3 grams

8 0

3 years ago

The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5

miskamm [114]

Answer:

z=-3.25

$p_v =2*P(z$

Step-by-step explanation:

1) Data given and notation

$\bar X=5.24$ represent the mean production for the sample

$\sigma=0.32$ represent the sample standard deviation for the sample

$n=16$ sample size

$\mu_o =5.5$ represent the value that we want to test

$\alph=0.05a$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean production is different from 5.5 tons, the system of hypothesis would be:

Null hypothesis: $\mu =5.5$

Alternative hypothesis: $\mu \neq 5.5$

If we analyze the info given we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{5.24-5.5}{\frac{0.32}{\sqrt{16}}}=-3.25$

P-value

Since this is a two side test the p value would be:

$p_v =2*P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the production its significant different compared to the desired percentage of SiO2 of 5.5 at 5% of signficance.

7 0

2 years ago

The length of the side of square A is 50% of the length of the square side expressing the area of the shaded area of square

KonstantinChe [14]

Answer:

Let's call the side of square b x. Therefore, the side of square a is 0.5x, making its area 0.25x² and the area of square b is x². This means that the percentage is 25%.

8 0

3 years ago

11. Identify the properties below.<br> a. 5 + 6 = 6 + 5<br> b. 7 + (3 + 9) = (7 + 9) + 3

jenyasd209 [6]

A is commutative property
B is associative property
Explanation:

5 0

3 years ago

Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.

masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is $g^{-1}(x) = e^{x}$

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is $g^{-1}(x) = e^{x}$

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y = ln(x)

Taking exponents of both sides, we have

$e^{y} = e^{lnx} \\e^{y} = x$

Replacing x with y, we have

$y = e^{x}$

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is $g^{-1}(x) = e^{x}$

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0

2 years ago