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RoseWind [281]
1 year ago
10

Is the following conditional true? If x < 4, then x < 10 O yes O no

Mathematics
1 answer:
Kay [80]1 year ago
7 0

Answer: No, i think so.

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Last one i need help with
LenKa [72]

Answer: 10 sq units

Step-by-step explanation:

3 + 4.5 + 0.5 + 2 = 10 sq units

6 0
3 years ago
What is the end behavior for f (x) = -3x4
Xelga [282]

Answer:

Step-by-step explanation:

If m is odd, that means the graph crosses the x-axis at x=a, if m is even, that means the graph touches the x-axis at x=a. x = a . The end-behavior of a polynomial function is given by the dominant term, anxn, a n x n , limx→±∞f(x)=limx→±∞anxn. lim x → ± ∞ f ( x ) = lim x → ± ∞ a n x n .  HERE HOPE THIS WORK.

5 0
4 years ago
2. 4x - 2y = -6<br> -6x + 2y = 2
Phoenix [80]

Answer:

x = 2, y =7

Step-by-step explanation:

4x - 2y = -6

-6x + 2y = 2

Add the equations together

4x - 2y = -6

-6x + 2y = 2

-----------------------

-2x = -4

Divide each side by -2

-2x/-2 = -4/-2

x = 2

now find y

-6x+2y =2

-6(2) +2y =2

-12+2y =2

Add 12 to each side

-12+12+2y = 2+12

2y =14

Divide by 2

2y/2 =14/2

y =7

6 0
4 years ago
A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \
Yanka [14]

Answer:

a) P(A)=0.875

b) \text{P(A or B)}=0.875

c) \text{P((not A)  or B  or (not C))}=0.625

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

P(A)=\frac{n(A)}{n(S)}

P(A)=\frac{7}{8}

P(A)=0.875

b) P(A or B)

Using formula,

\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}

\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}

\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}

\text{P(A or B)}=\frac{7}{8}

\text{P(A or B)}=0.875

(c) P((not A)  or B  or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

\text{P((not A)  or B  or (not C))}=\frac{5}{8}

\text{P((not A)  or B  or (not C))}=0.625

4 0
3 years ago
How would I show twelve in four different ways? It said to use words, pictures, or numbers
vesna_86 [32]

Answer:

twelve, 12 and just draw twelve dots

8 0
3 years ago
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