Add this fraction 6 3/8 4 1/8

1 answer:

8 0

$6\frac{3}{8} + 4\frac{1}{8}$

1. Add the whole numbers

$6 + 4 = 10$

2. Add the fractions

$\frac{3}{8} + \frac{1}{8} = \frac{4}{8}$

$6\frac{3}{8} + 4\frac{1}{8} = 10\frac{4}{8}$

3. Reduce

$10\frac{4 / 4}{8 / 4} = 10\frac{1}{2}$

$6\frac{3}{8} + 4\frac{1}{8} = 10\frac{4}{8} = 10\frac{1}{2}$

The answer is 10 1/2

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(x + 4) (x - 4) - x² + 16

Nitella [24]

Answer:

infinite solutions

Step-by-step explanation:

Hey there!

In order to solve this problem, we need to first distribute the $(x+4)(x-4)$

This is equal to $x^2-16$

Now we can combine like terms

$x^2$ and $-x^2$ cancel out

also

$16$ and $-16$ cancel out

The equation ends up as 0 = 0

So the answer to this answer is infinite solutions

6 0

2 years ago

According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 Assume a

Veseljchak [2.6K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, $\mu$ = 2.58 and Standard deviation, $\sigma$ = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar $\leq$ 2.50)

P(Xbar < 2.66) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z < -1.68) = 1 - P(Z 1.68)

= 1 - 0.95352 = 0.04648

P(Xbar $\leq$ 2.50) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }$ ) = P(Z $\leq$ -3.92) = 1 - P(Z < 3.92)