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scoray [572]
3 years ago
7

Please help! and explain thank you!

Mathematics
2 answers:
ExtremeBDS [4]3 years ago
6 0

Answer:

D

Step-by-step explanation:

You want to move all terms not containing a to the right side of the equation.

Tju [1.3M]3 years ago
4 0

Answer:

B

Step-by-step explanation:

When you are solving these kinds of problems, it's always best to start with getting rid of the decimal/fraction.

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

We can actually solve for a.

2/3*3=6/3=2

(15+2a=-5)

a=-10

(15+(-20)=-5)

TRUE

a=10 (I think)

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Radda [10]

Answer:

what?

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Ion
erik [133]

Answer: It is being dilated by a scale factor of 1/3.

Step-by-step explanation:

The pre-image has the coordinates (0,0) (6,0) (6,-3) and (0,-3) and a transformation has been applied to it to get the coordinates of the image to be (0,0) (2,0) (2,-1) ( 0,-1)

Looking at the coordinates we can see that to get from 6 to 2 you will have to multiply by  1/3 and to  get from -3 to -1 you will have to multiply by 1/3.

This means that pre-image was dilated  by a scale factor of 1/3.

8 0
4 years ago
Assume that when adults with smartphones are randomly​ selected, 51​% use them in meetings or classes. If 11 adult smartphone us
Luba_88 [7]

Answer:

The probability is 0.2356.

Step-by-step explanation:

Let X is the event of using the smartphone in meetings or classes,

Given,

The probability of using the smartphone in meetings or classes, p = 51 % = 0.51,

So, the probability of not using smartphone in meetings or classes, q = 1 - p = 1 - 0.51 = 0.49,

Thus, the probability that fewer than 5 of them use their smartphones in meetings or classes.

P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3)+P(X=4)

Since, binomial distribution formula is,

P(x) = ^nC_r p^x q^{n-x}

Where, ^nC_r=\frac{n!}{r!(n-r)!}

Here, n = 11,

Hence,  the probability that fewer than 5 of them use their smartphones in meetings or classes

=^{11}C_0 (0.5)^0 0.49^{11}+^{11}C_1 (0.5)^1 0.49^{10}+^{11}C_2 (0.5)^2 0.49^{9}+^{11}C_3 (0.5)^3 0.49^{8}+^{11}C_4 (0.5)^4 0.49^{7}

=(0.5)^0 0.49^{11}+11(0.5)0.49^{10} + 55(0.5)^2 0.49^{9}+165 (0.5)^3 0.49^{8} +330(0.5)^4 0.49^{7}

=0.235596671797

\approx 0.2356

4 0
4 years ago
For each of the following sequences, • Give a formula for the nth term in the sequence, • Give a recursive definition for the se
Andrew [12]

Answer:

(a)n^{th} = n

f(1) = 1

f(n) = f(n-1) + 1

(b)n^{th} = 2^{n-1}

f(1) = 1

f(n) = f(n-1) * 2

(c)n^{th} = n!

f(1) = 1

f(n) = f(n-1) * n

Step-by-step explanation:

(a) This is a sequence of consecutive number

n^{th} = n

f(1) = 1

f(n) = f(n-1) + 1

(b) This is a sequence of 2 to the power of n - 1. The next number is twice time of this number

n^{th} = 2^{n-1}

f(1) = 1

f(n) = f(n-1) * 2

(c) This is factorial sequence. Where the next number is this number multiplied by n^{th}

n^{th} = n!

f(1) = 1

f(n) = f(n-1) * n

5 0
3 years ago
What is the equation of the following line? Be sure to scroll down first to see all answer options (2,10) (0,0)
nordsb [41]
Y=5x is the correct answer
7 0
3 years ago
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