64^-3/2*1331^-1/3 evaluate

A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the

olga nikolaevna [1]

Answer: Option (A)

Step-by-step explanation:

Let the original price of the clock be "x "

The clock is then resold to a collector = x + 20% of "x "

= x + 0.2x

= 1.2x

Now, the shop re-buy the clock at a price = $Reselling\ price \times 0.50$

$= 1.2x \times 0.5$

= 0.6x

Therefore,

Original Cost - Buy-Back price = $100

⇒ x - 0.6x = 100

⇒ x = 250

Hence , the selling price of the clock for the second time is

$= 1.8\times0.6\times250$

= $270

7 0

3 years ago

HELP. I NEEEED HELP. THANK YOU.

rodikova [14]

Answer:

The slope is 1/2

Step-by-step explanation:

If you watch where each dot landes, it's half of each y number. The sodas cost $0.50 which is half of 1.

6 0

2 years ago

Read 2 more answers

Solve the proportion below.

melamori03 [73]

Cross multiply

19(x) = 30(9.5)

Simplify

19x = 285

Divide 19 from both sides

19x/19 = 285/19

x = 285/19

x = 15

B. 15 is your answer

hope this helps

8 0

2 years ago

Read 2 more answers

A physicist examines 25 water samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.165 cc/

xz_007 [3.2K]

Answer:

The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).

Critical value t=1.318

Step-by-step explanation:

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=0.165.

The sample size is N=25.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.078}{\sqrt{25}}=\dfrac{0.078}{5}=0.016$

The degrees of freedom for this sample size are:

$df=n-1=25-1=24$

The t-value for a 80% confidence interval and 24 degrees of freedom is t=1.318.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.318 \cdot 0.016=0.021$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 0.165-0.021=0.144\\\\UL=M+t \cdot s_M = 0.165+0.021=0.186$

The 80% confidence interval for the population mean nitrate concentration is (0.144, 0.186).

6 0

3 years ago

g A coffee-dispensing machine is supposed to deliver eight ounces of liquid into each paper cup, but a consumer believes that th

kap26 [50]

Answer: Reject the eight- ounces claim.

Step-by-step explanation:

For left tailed test , On a normal curve the rejection area lies on the left side of the critical value.

It means that if the observed z-value is less than the critical value then it will fall into the rejection region other wise not.

As per given ,

Objective : A coffee-dispensing machine is supposed to deliver eight ounces of liquid or less.

Then ,

$H_0: \mu=8\\\\H_a: \mu$ , since alternative hypothesis is left-tailed thus the test is an left-tailed test.

the critical value for z for a one-tailed test with the tail in the left end is -1.645 and the obtained value is -1.87.

Clearly , -1.87 < -1.645

⇒ -1.87 falls under rejection region.

⇒ Decision : Reject null hypothesis.

i.e. we reject the eight- ounces claim.

6 0

3 years ago