What does this mean ? cos θ

Katena32 [7]

Answer:

Just get the perimeter of the smaller rectangle then times it by 4

Step-by-step explanation:

i tried to use a calculator but i cant do the perimeter of the smaller triangle because i dont know what x is sorry ill try to figure it out.

8 0

3 years ago

What is the lowest value of the range of the function

Rasek [7]

The question is incomplete. Below you will find the missing graph.

The lowest value of the range of the function shown on graph is (B) -2.

Graph is the pictorial representation of the function or mathematical relation.

The lowest value of the graph is the value of y when $y=f(x)$ gives the least value.

From the given graph we can clearly see that at $x=3$ the graph of the given function approaches the least value which is -2.

So the lowest value of the range of the function shown on the graph is -2.

Hence the correct option is (B).

Learn more about Graph here -

brainly.com/question/4025726

#SPJ10

7 0

2 years ago

Please help with this will give brainliest to the first answer!

kozerog [31]

Answer:

your answer is 415 in^3

Step-by-step explanation:

ok the equation in finding the volume of a come is

V= $\pi r^2 \frac{h}{3}$

step 1: plugin

$\pi =3.14$

3.14 x 6^2 x 11/3

3.14x 36 x 11/3 = (6/3 turns into a 2)

113.04x 11/3= 114. 48, when rounded, it's going to give you 415 in^3

5 0

2 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

g A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is cha

leva [86]

Answer:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

$z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502$

$p_v =2*P(Z>2.502)= 0.0123$

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.

Step-by-step explanation:

Data given and notation

$X_{1}=410$ represent the number of people indicating that their financial security was more than fair for the recent year

$X_{2}=245$ represent the number of people indicating that their financial security was more than fair for the year before

$n_{1}=1000$ sample 1 selected

$n_{2}=700$ sample 2 selected

$p_{1}=\frac{410}{1000}=0.410$ represent the proportion estimated of indicating that their financial security was more than fair this year

$p_{2}=\frac{245}{700}=0.35$ represent the proportion estimated of indicating that their financial security was more than fair the year before

$\hat p$ represent the pooled estimate of p

z would represent the statistic (variable of interest)

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

$\alpha$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{410+245}{1000+700}=0.385$

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502$

Statistical decision

Since is a two sided test the p value would be:

$p_v =2*P(Z>2.502)= 0.0123$

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.

7 0

3 years ago

What does this mean ? cos θ > 0