8x + 2y = 44
5x + 2y = 35

Since 2y is already common, there is no need for multiplication. Just reverse the signs on the equation.

Like this;

8x + 2y = 44
-5x - 2y = -35

All the additions become subtractions and vice versa.

-2y and +2y get cancelled.

Leaving;

8x = 44
-5x = -35

Now, subtract again.

8x - 5x = 3x

44 - 35 = 9

So;

3x = 9

x = 9 / 3

x = 3

We found 'x' and now we have to find 'y'.

Just substitute.

[8 x 3] + 2y = 44

24 + 2y = 44

2y = 44 - 24

2y = 20

y = 20 / 2

y = 10

Now, we re-check;

[8 x 3] + [2 x 10] = 44

24 + 20 = 44

Explaining the error for #9b is simply doing the whole sum.

Hope this helped! :) I tried my best to explain.

7 0

3 years ago

At 1 p.m., the number of cars in a parking lot is 78. Given the formula found in the first part of the question, An=78⋅1.09n, es

Harrizon [31]

Answer:

At 6 p.m., the number of cars in a parking lot is 120

Step-by-step explanation:

At 1 p.m., the number of cars in a parking lot is 78.

Given,

$A_n=78\cdot 1.09^n$

Note that

$A_0=78\cdot 1.09^0=78\cdot 1=78$

So, 1 p.m. goes for n = 0, then

6 p.m. goes for n = 5

Thus,

$A_5=78\cdot 1.09^5\approx 120$

4 0

3 years ago

A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken

mihalych1998 [28]

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

$X=1$ number of citrus trees that are infected with the citrus red mite.

$n=5$ random sample taken

$\hat p=\frac{1}{5}=0.2$ estimated proportion of citrus trees that are infected with the citrus red mite.

$p$ true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical values would be given by: