Number line from negative 10 to 10 , tic marks at every integer , labels at negative 10 , negative 5 , 0 , 5 , 10 , closed red circle on negative 1 , red shading to the right of the red circle. A.
y + 5 ≤ 4

B.
y + 5 < 4

C.
y + 5 ≥ 4

D.
y + 5 > 4

Answers

Answer 1

Answer: The right answer for the question that is being asked and shown above is that: "B. y + 5 < 4"Number line from negative 10 to 10 , tic marks at every integer , labels at negative 10 , negative 5 , 0 , 5 , 10 , closed red circle on negative 1 , red shading to the right of the red circle.

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The number of members in JJ Select in 1980 was approximately 2 × 102 people. By 2005, the number of members had risen to 6 × 105 people. Which of the following is true?

Given: 5x + 3 > 4x + 7. Choose the graph of the solution set.

Answers

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$5x+3 > 4x+7$

solve for x

Group terms that contain the same variable, and move the constant to the opposite side of the inequality

$5x-4x > 7-3$

$x> 4$

The solution is the interval-------> (4,∞)

All real numbers greater than $4$

The solution is the shaded area to the right of the vertical dashed line

see the attached figure

Let's solve your inequality step-by-step.

5x+3>4x+7

Step 1: Subtract 4x from both sides.

5x+3−4x>4x+7−4x

x+3>7

Step 2: Subtract 3 from both sides.

x+3−3>7−3

x>4

Answer:

x>4

Which relation is a function?

Answers

Of the provided graphs, the second would be the correct answer.

Functions occur when the input only has one possible output (though the output can be recieved through multiple inputs)

The relation shown in the 2nd diagram is a function. The reason for this is that one and only one y-value is associated with each x-value.

Note that this is not true of the first diagram. If x=0, there are two separate y vaues associated: -1 and 2. This clearly states that the first diagram does not represent a function.

John is playing a game of darts. The probability that he throws a dart into the center of the dart board (the Bull’s eye) is 1/10. The probability that he throws the dart into the 10-point ring is 3/10.What is the probability that he either hits a Bull’s eye or scores 10 points?
a. 1/3
b. 2/3
c. 3/5
d. 2/5
e. 1/4

Answers

The correct answer for the question that is being presented above is this one: "d. 2/5." John is playing a game of darts. The probability that he throws a dart into the center of the dart board (the Bull’s eye) is 1/10. The probability that he throws the dart into the 10-point ring is 3/10. The probability that he either hits a Bull's eye or scores 10 points is 2/5

Answer: The correct option is (d). $(2)/(5).$

Step-by-step explanation: Given that John is playing a game of darts. The probability that he throws a dart into the centre of the dart board (the Bull’s eye) is $(1)/(10)$ and the probability that he throws the dart into the 10-point ring is $(3)/(10).$

We are to find the probability that he either hits a Bull’s eye or scores 10 points.

Let, 'A' and 'B' represents the events that John throws the dart into a Bull's eye and 10-point ring respectively.

Then, according to the given information, we have

$P(A)=(1)/(10),~~P(B)=(3)/(10),~~~P(A\cup B)=?$

Since John cannot throw the dart into the Bull's eye and 10 point ring together, both the events are independent of each other.

Therefore,

$P(A\cap B)=0$

From the theorems of probability, we have

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=(1)/(10)+(3)/(10)-0=(4)/(10)=(2)/(5).$

Therefore, the probability that John either hits a Bull’s eye or scores 10 points is $(2)/(5).$

Thus, (d) is the correct option.

17.21 million without dicemal

Answers

17.21 million without dicemal

Ans= 17,210,000

Two angles are complementary. If one angle measures 25 degrees, what is the measure of the second angle?A. 65 degrees
B. 75 degrees
C. 155 degrees
D. 165 degrees

Answers

Answer:

i think its C also

Step-by-step explanation:

I believe it is c.155 degrees

Find the fifth term of the arithmetic sequence in which t1 = 3 and tn = tn-1 + 4.

A) 5
B) 7
C) 19
D) 23

Answers

We have given $t_(1)=3$ and $t_(n) = t_(n-1) + 4$ . Fifth term is $t_(5)= t_(4)+4$ . It means we need to find $t_(4)$ and in order to find the latter, we have to find $t_(4)= t_(3)+4$ , $t_(3)= t_(2)+4$ and $t_(2)= t_(1)+4$ . Since we know the value of $t_(1)$ , beginning from the last equation we can find the value of $t_(5)$ . We can write that $t_(2)= t_(1)+4=3+4=7$ , $t_(3)= t_(2)+4=7+4=11$ and $t_(4)= t_(3)+4=11+4=15$ . Since $t_(5)= t_(4)+4$ , then $t_(5)= 15+4=19$

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Number line from negative 10 to 10 , tic marks at every integer , labels at negative 10 , negative 5 , 0 , 5 , 10 , closed red circle on negative 1 , red shading to the right of the red circle. A. y + 5 ≤ 4 B. y + 5 < 4 C. y + 5 ≥ 4 D. y + 5 > 4

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Number line from negative 10 to 10 , tic marks at every integer , labels at negative 10 , negative 5 , 0 , 5 , 10 , closed red circle on negative 1 , red shading to the right of the red circle. A.
y + 5 ≤ 4

B.
y + 5 < 4

C.
y + 5 ≥ 4

D.
y + 5 > 4