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3 years ago

Write the intersection and union of N and W.

Mathematics

1 answer:

Serjik [45]3 years ago

5 0

Answer:

N n W = {1,2,3,4}

N u W = {0,1,2,3,4,}

Step-by-step explanation:

N n W = {1,2,3,4}

N u W = {0,1,2,3,4,}

n - intersection means the common elements between N and W

U - represents all elements within N and W

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Probability of rolling exactly 1 odd number when 3 6-sided dice are rolled

denis-greek [22]

Answer:

9/18

Step-by-step explanation:

4 0

2 years ago

What is the 52nd term of the sequence below?

VladimirAG [237]

Because the difference between any term and the previous term is a constant, this is an arithmetic sequence because of that constant which is referred to as the common difference, d. Which in this case is -35--38=-32--35=3

Any arithmetic sequence can be expressed as:

a(n)=a+d(n-1), a=initial term, d=common difference, n=term number.

In this case a=-38 and d=3 so

a(n)=-38+3(n-1) which we can simplify to

a(n)=-38+3n-3

a(n)=3n-41, so the 52nd term is:

a(52)=3(52)-41

a(52)=156-41

a(52)=115

3 0

3 years ago

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

3 years ago

Evaluate the function when f(x) = 2x + 2, when x = -1.

maks197457 [2]

Answer: 2 x -1 = -2. -2 + 2 =0

6 0

3 years ago

Read 2 more answers

The equation of the piecewise function f(x) is below. What is the value of f(3)?

JulijaS [17]

Option B: 5 is the value of f(3)

Explanation:

The equation of the piecewise function is given by

$f(x)=\left\{\begin{aligned}-x^{2}, & x$

We need to find the value of $f(3)$

The value of the function f can be determined when $x=3$ by identifying in which interval does the value of $x=3$ lie in the piecewise function.

Thus, $x=3$ lies in the interval $x\geq 0$ , the function f is given by

$f(x)=x+2$

Substituting $x=3$ in the function $f(x)=x+2$ , we get,

$f(3)=3+2$

$f(3)=5$

Thus, the value of $f(3)$ is 5.

Therefore, Option B is the correct answer.

8 0

3 years ago