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3 years ago

Please help ill brainlist and put higher points in my next question

Mathematics

1 answer:

Bumek [7]3 years ago

4 0

Answer:

Positive

Step-by-step explanation:

Since b is already a negative, and b is greater than a, that means a will be a positive.

example:

Imagine a = -4

4 x -4 = -16

Imagine b = -5

-3 x -5 = 15

-4 + 15 = 11, which is positive

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A- y= -3/2x+8<br> B- y=3/2x+5<br> C- y=2/3x+8<br> D- y=-2/3x+5

Annette [7]

Answer:

I think the right answer is B.

4 0

3 years ago

B<br> What does a single point on the plot represent?

NemiM [27]

It represents a function

6 0

3 years ago

Read 2 more answers

If all possible samples of size n are drawn from an infinite population with a mean of 20 and a standard deviation of 5, then th

Ugo [173]

Answer:

Step-by-step explanation:

We know that the standard error of the sampling distribution of sample means is

Standard error=standard deviation/√n

√n=standard deviation/Standard error

n=(standard deviation/Standard error)²

We are given that standard deviation=5 and standard error=1. So,

n=(5/1)²

n=5²

n=25.

Thus, the required sample size is 25.

3 0

3 years ago

Write a division expression that could also be solved by using 1/2x9?

aleksandrvk [35]

Answer:

18 divided by 4.

Step-by-step explanation:

1/2x9= 4 1/2 and when you divide 18 divided by 4 you get 4 1/2 (4.5)

8 0

3 years ago

Question 3 (1 point) Co-60 has a half life of 5.3 years. If a pellet that has been in storage for 26.5 years contains 14.5g of C

Zarrin [17]

Answer:

464 grams.

Step-by-step explanation:

Amount of substance:

The amount of substance after t years is given by an equation in the following format:

$A(t) = A(0)(1-r)^t$

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Co-60 has a half life of 5.3 years.

This means that:

$A(5.3) = 0.5A(0)$

We use this to find r. So

$A(t) = A(0)(1-r)^t$

$0.5A(0) = A(0)(1-r)^{5.3}$

$(1-r)^{5.3} = 0.5$

$\sqrt[5.3]{(1-r)^{5.3}} = \sqrt[5.3]{0.5}$

$1 - r = 0.5^{\frac{1}{5.3}}$

$1 - r = 0.8774$

$A(t) = A(0)(0.8774)^{t}$

If a pellet that has been in storage for 26.5 years contains 14.5g of Co-60, how much of this radioisotope was present when the pellet was put in storage?

We have that $A(26.5) = 14.5$ , and use this to find A(0). So

$A(t) = A(0)(0.8774)^{t}$

$14.5 = A(0)(0.8774)^{26.5}$

$A(0) = \frac{14.5}{(0.8774)^{26.5}}$

[tex]A(0) = 464[tex]

464 grams.

3 0

3 years ago