The sum of one half and one third, halved 5th grade explation

Economists predict that Americans will spend $1,180 on Summer Vacation in 2015, this is up 5.6% from 2014. What did Americans sp

netineya [11]

Answer: $1117.42

Step-by-step explanation:

Given: The amount predicted by Economist =$1180

Let x be the amount spent by Americans on Summer vacation in 2014.

Therefore, According to the question, we have

5.6% of x+x=1180

$\Rightarrow0.056\timesx+x=1180\\\Rightarrow x(0.056+1)=1180\\\Rightarrow\ x=\frac{1180}{1.056}=1117.42\approx$

Hence, the amount spent by Americans on Summer vacation in 2014 = $1117.42

3 0

3 years ago

2 divion 18 I need helpppppppp

Gnoma [55]

1 divided by 18 is .111111111

3 0

3 years ago

Read 2 more answers

Mark’s Pool Repair charges $50 for a service call, plus $40/h for labour. Jennifer’s Pools and Spas charges $30 for a service ca

tatyana61 [14]

Multiply each hourly rate by x ( time to complete the work) add it to the service call for each and then set the equations equal:

50 + 40x = 30 +45x

Subtract 30 from both sides:

20 + 40x = 45x

Subtract 40x from both sides:

20 = 5x

Divide both sides by 5

x = 4

The length is 4 hours.

3 0

3 years ago

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally

rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.

So, X ~ Normal( $\mu=5.0,\sigma^{2} =1.7^{2}$ )

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = population mean thickness = 5.0 mm

$\sigma$ = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

P(X < 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{3.0-5.0}{1.7}$ ) = P(Z < -1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

P(X > 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ > $\frac{7.0-5.0}{1.7}$ ) = P(Z > 1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X $\leq$ 3.0 mm)

P(X < 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{7.0-5.0}{1.7}$ ) = P(Z < 1.18) = 0.881

P(X $\leq$ 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{3.0-5.0}{1.7}$ ) = P(Z $\leq$ -1.18) = 1 - P(Z < 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

4 0

3 years ago

What is the answer to this problem 6(1-5m)

Ivahew [28]

Apply the distributive property.6⋅1+6(−5m)

Multiply 66 by 11 to get 66.6+6(−5m)6+6(-5m)Multiply −5-5 by 66 to get −30-30.6−30m6-30mReorder 66 and −30m-30m.−30m+6
So your answer is −30m+6

5 0

3 years ago