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3 years ago

PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER!!

Mathematics

1 answer:

PtichkaEL [24]3 years ago

6 0

Answer:

Step-by-step explanation:

Let's go over each option with the process of elimination!

A. A data set will always have exactly one mode.

Contradiction: 2, 2, 3, 3, 4, 5, 6

in this data set, both 2 and 3 are modes, so it is incorrect.

B. A data set can have multiple means.

True, example as A.

C. A data set will have at least one mode.

Contradiction: 2, 3, 4, 5, 6

There are no modes in here.

D. A data set will never have a mode.

Contradiction: 2, 2, 3, 4, 5

The mode is 2.

The only true statement is answer B.

I hope this helped! :D

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Determine whether each equation is True or False. In case you find a "False" equation, explain why is False.

elixir [45]

Answer:

(1) TRUE.

(2) FALSE.

(3) FALSE.

(4) TRUE.

(5) FALSE.

Step-by-step explanation:

(1) $\sqrt{32} = 2^{\frac{5}{2} }$

$2^{\frac{5}{2} } = (\sqrt{2} )^5 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 4\sqrt{2}\\\\\sqrt{32} = \sqrt{16 \ \times \ 2}\ = \ \sqrt{16} \ \times \ \sqrt{2} \ = \ 4\sqrt{2}$

Thus, the equation is TRUE.

(2) $16^{\frac{3}{8} } = 8^2$

$16^{\frac{3}{8} } =(2^4)^{\frac{3}{8} } = 2^\frac{3}{2} }= (\sqrt{2} )^3 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 2\sqrt{2} \\\\8^2 = 64$

Thus, the equation is FALSE.

(3) $4^{\frac{1}{2} } = \sqrt[4]{64}$

$4^{\frac{1}{2} }= \sqrt{4} = 2\\\\\sqrt[4]{64} = (64)^{\frac{1}{4} } = (2^6)^{\frac{1}{4} }= 2^{\frac{6}{4} } = 2^{\frac{3}{2} }=(\sqrt{2} )^3 = (\sqrt{2} \times \sqrt{2} \times \sqrt{2} ) = 2\sqrt{2}$

Thus, the equation is FALSE.

(4) $2^8 = (\sqrt[3]{16} )^6$

$2^8 = 256\\\\ (\sqrt[3]{16} )^6 = (16)^{\frac{6}{3} } = (2^4)^{\frac{6}{3} } = (2)^{\frac{24}{3} } = 2^8 = 256$

Thus, the equation is TRUE.

(5) $(\sqrt{64} )^{\frac{1}{3} } = 8^{\frac{1}{6} }\\\\$

$8^{\frac{1}{6} } = (2^3)^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} } = \sqrt{2} \\\\(\sqrt{64} )^{\frac{1}{3} } = (2^6)^{\frac{1}{3} } = 2^{\frac{6}{3} } = 2^2 = 4$

Thus, the equation is FALSE.

4 0

2 years ago

What is the range of the relation? {(1, 2), (2, 4), (3, 2), (4, 6)} A.{2, 4} B.{1, 2, 3, 4, 6} C.{2, 4, 6} D.{1, 2, 3, 4}

tamaranim1 [39]

$(1)\ \ \ \ range:\ \ \ \{2, 4, 6\}\ \ \ \Rightarrow\ \ \ Ans.\ C\\\\(2)\ \ \ function:\ \ \ \{(1, 2); (2, 2); (3, 2); (4, 2)\}\ \ \ \Rightarrow\ \ \ Ans.\ D\\\\(3)\ \ \ f(4)=2-3\cdot4=2-12=-10\ \ \ \Rightarrow\ \ \ Ans.\ C\\\\(4)\ \ \ direct:\ \ \ y=ax\ \ \ \Rightarrow\ \ \ y= \frac{1}{3} x\ \ \ \Rightarrow\ \ \ Ans.\ B\\\\(5)\ \ \ f(x)=ax\ \ \ and\ \ \ f(4)=12\\.\ \ \ \ \ \Rightarrow\ \ \ 12=a\cdot4\ \ \ \Rightarrow\ \ \ a=3\ \ \ \Rightarrow\ \ \ f(x)=3x\ \ \ \Rightarrow\ \ \ Ans.\ (?)\\\\$

$(6)\ \ \ f(x)=ax^2\ \ \ and\ \ \ f(2)=48\\.\ \ \ \Rightarrow\ \ 48=a\cdot2^2\ \ \Rightarrow\ \ a=48:4=12\ \ \Rightarrow\ \ f(x)=12x^2\ \Rightarrow\ \ Ans.\ C\\\\(7)\ \ \ inversely:\ \ \ f(x)= \frac{a}{x} \ \ \ and\ \ \ f(2)=4\\.\ \ \ \Rightarrow\ \ \ 4= \frac{a}{2} \ \ \ \Rightarrow\ \ \ a=4\cdot2=8\ \ \ \Rightarrow\ \ \ y= \frac{8}{x}\ \ \ \Rightarrow\ \ \ Ans.\ B$

5 0

3 years ago

HELP ME I AM CRYING SO MUCH ALSO LMK IF U CANT SEE

Wittaler [7]

Send a better picture so I can help.

3 0

3 years ago

A parallel system functions whenever at least one of its components works. Consider a parallel system of 5 components, and suppo

Aneli [31]

Answer:

Probability that component 4 works given that the system is functioning = 0.434 .

Step-by-step explanation:

We are given that a parallel system functions whenever at least one of its components works.

There are parallel system of 5 components and each component works independently with probability 0.4 .

Let A = Probability of component 4 working properly, P(A) = 0.4 .

Also let S = Probability that system is functioning for whole 5 components, P(S)

Now, the conditional probability that component 4 works given that the system is functioning is given by P(A/S) ;

P(A/S) = {Means P(component 4 working and system also working)

divided by P(system is functioning)}

P(A/S) = {In numerator it is P(component 4 working) and in

denominator it is P(system working) = 1 - P(system is not working)}

Since we know that P(system not working) means that none of the components is working in system and it is given with the probability of 0.6 and since there are total of 5 components so P(system working) = 1 - $0.6^{5}$ .