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katen-ka-za [31]

3 years ago

5

Complete the following table so that it represents a quadratic function.

2 answers:

Softa [21]3 years ago

5 0

You have multiple posted i just did this on hr other one lol

Annette [7]3 years ago

4 0

I don’t know honey con found dirty gm dry cod CMC gold cam bring all can

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The nicotine content in cigarettes of a certain brand is normally distributed with mean

natita [175]

Hi am I am in my car now I’m playing with my car I will

4 0

3 years ago

A closed rectangular box has volume 38 cm ^3. What are the lengths of the edges giving the minimum surface area

Margaret [11]

Answer:

3.361 3.361 3.361 cm

Step-by-step explanation:

Minimum surface are will be a cube

LWH = 38

L = cubrt (38) = 3.361 side length cm

8 0

3 years ago

1. Suppose D has a coordinate of 11 and DE = 5. What are the possible<br> midpoints of DE?

Rasek [7]

Answer:

The possible midpoints are (5.5,8) and (5.5,3)

Step-by-step explanation:

Given

$D = 11$

$DE = 5$

Required

Determine the midpoint of DE

First, we need to determine the coordinates of E

If E is to the right of DE;

$E - D = 5$

$E - 11 = 5$

$E = 11 + 5$

$E = 16$

Hence, the coordinate is (11,16)

$Midpoint = \frac{1}{2} * (11,16)$

$Midpoint = (5.5,8)$

Else:

$D - E = 5$

$11 - E = 5$

$E = 11 - 5$

$E = 6$

Hence, the coordinate is (11,6)

$Midpoint = \frac{1}{2} * (11,6)$

$Midpoint = (5.5,3)$

<em>The possible midpoints are (5.5,8) and (5.5,3)</em>

6 0

3 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago

I help unit circle questions Help please!

Paha777 [63]

What is the quest? It doesn’t show it

5 0

3 years ago