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nikitadnepr [17]

3 years ago

11

Determine whether a triangle can be formed with the given side lengths.

1 answer:

iren [92.7K]3 years ago

4 0

Answer:

A triangle can have these three side lengths.

You might be interested in

Solve for x.3-sqrt(x)=0 What is the root? If there is none, choose none. x = 3 x = 9 none

Anon25 [30]

Answer:

x = 9

Step-by-step explanation:

You can solve this problem by isolating the variable.

3-√x = 0

√x = 3 (add √x to both sides of the equation)

x = 9 (square both sides of the equation)

8 0

3 years ago

How many fluid ounces in 8 ounces and 5 cups?

xenn [34]

<span>48 fluid ounces is the right answer</span>

3 0

4 years ago

Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypot

Nina [5.8K]

Answer: a. 0.1031; fail to reject the null hypothesis

Step-by-step explanation:

Given: Significance level : $\alpha=0.05$

The test statistic in a two-tailed test is z = -1.63.

The P-value for two-tailed test : $2P(Z>|z|)=2P(Z>|-1.63|)=0.1031$ [By p-value table]

Since, 0.1031 > 0.05

i.e. p-value > $\alpha$

So, we fail to reject the null hypothesis. [When p< $\alpha$ then we reject null hypothesis ]

So, the correct option is a. 0.1031; fail to reject the null hypothesis.

8 0

3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

What is the surface area of a rectangular prism that has a length of 9 feet a width of 5 feet and a height of 7.5 feet

lukranit [14]

Answer:

300

Step-by-step explanation:

find the sum of area of all sides

6 0

3 years ago