What is the answer to this problem? 2 1/3 + 3 1/3 = ?

tekilochka [14]

Since the exponent (-3) is negative, flip the expression to: 1/(5n^4)^3.

Notice how the negative exponent becomes positive as you flip it.

Now evaluate the powers in the denominator: 1/((5^3)(n^4)^3

I separated the constant (5) from the variable (n) to show you how the powers are evaluated.

1/(5x5x5)(nxnxnxn)(nxnxnxn)(nxnxnxn)
—-> the power four means that there are 4 multiples of n in the parentheses. The power 3 corresponds to how many groups.

1/(125)(n^12)

= 1/125n^12

6 0

3 years ago

Create 3 angle that will create a triangle. Remember, a triangle has to have a sum of 180 degrees. I then want you to respond ba

levacccp [35]

Answer:

60,60,60

90, 45,45

Step-by-step explanation:

I'll name them Jim and Jam

:)

5 0

3 years ago

Hats and mittens are on sale at the store! One woman was able to buy 5 hats and 4 pairs of mittens for $30.

morpeh [17]

The price of one hat is $2 and price of one pair of mittens is $5.

Step-by-step explanation:

Let,

Price of one hat = x

Price of one pair of mittens = y

According to given statement;

5x+4y=30 Eqn 1

2x+3y=19 Eqn 2

Multiplying Eqn 1 by 2

$2(5x+4y=30)\\10x+8y=60\ \ \ \ Eqn\ 3$

Multiplying Eqn 2 by 5

$5(2x+3y=19)\\10x+15y=95\ \ \ \ Eqn\ 4$

Subtracting Eqn 3 from Eqn 4

$(10x+15y)-(10x+8y)=95-60\\10x+15y-10x-8y=35\\7y=35\\$

Dividing both sides by 7

$\frac{7y}{7}=\frac{35}{7}\\y=5$

Putting y=5 in Eqn 1

$5x+4(5)=30\\5x+20=30\\5x=30-20\\5x=10$

Dividing both sides by 5

$\frac{5x}{5}=\frac{10}{2}\\x=2$

The price of one hat is $2 and price of one pair of mittens is $5.

Keywords: linear equation, subtraction

Learn more about linear equations at:

#LearnwithBrainly

4 0

3 years ago

Jamie had 4 fith of a spool of twin.He then used half of a spool of twin to make freindship knots.Heclaims to have 3 tenth of th

Rom4ik [11]

Jamie's claim isn't reasonable. This is because half of 4/5 is 4/10 and if you simplify it, it is 2/5.

6 0

3 years ago

The mean of a population is 74 and the standard deviation is 16. The shape of the population is unknown. Determine the probabili

uysha [10]

Answer:

(a) P( $\bar X$ $\geq$ 76) = 0.2327

(b) P(73 < $\bar X$ < 75) = 0.5035

(c) P( $\bar X$ < 74.8) = 0.77035

Step-by-step explanation:

We are given that the mean of a population is 74 and the standard deviation is 16.

Assuming the data follows normal distribution.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

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

where, $\mu$ = population mean = 74

$\sigma$ = standard deviation = 16

n = sample size

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

(a) Probability that a random sample of size 34 yielding a sample mean of 76 or more is given by = P( $\bar X$ $\geq$ 76)

P( $\bar X$ $\geq$ 76) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{76-74}{\frac{16}{\sqrt{34} } }$ ) = P(Z $\geq$ 0.73) = 1 - P(Z < 0.73)

= 1 - 0.7673 = 0.2327

The above probability is calculated by looking at the value of x = 0.73 in the z table which has an area of 0.7673.

(b) Probability that a random sample of size 120 yielding a sample mean of between 73 and 75 is given by = P(73 < $\bar X$ < 75) = P( $\bar X$ < 75) - P( $\bar X$ $\leq$ 73)

P( $\bar X$ < 75) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{75-74}{\frac{16}{\sqrt{120} } }$ ) = P(Z < 0.68) = 0.75175

P( $\bar X$ $\leq$ 73) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{73-74}{\frac{16}{\sqrt{120} } }$ ) = P(Z $\leq$ -0.68) =1 - P(Z < 0.68)

= 1 - 0.75175 = 0.24825

Therefore, P(73 < $\bar X$ < 75) = 0.75175 - 0.24825 = 0.5035

The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.75175.

(c) Probability that a random sample of size 218 yielding a sample mean of less than 74.8 is given by = P( $\bar X$ < 74.8)

P( $\bar X$ < 74.8) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{74.8-74}{\frac{16}{\sqrt{218} } }$ ) = P(Z < 0.74) = 0.77035

The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.

7 0

3 years ago