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3 years ago

Decrease 342g by 8% plez

Mathematics

1 answer:

Tom [10]3 years ago

6 0

Just multiply 342g by .08 which is 8 % and you get 27.36
So then you decrease that 8% from 342

342-27.36=314.64

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Can someone please help me

Alina [70]

Yes with what ? are u single ?

5 0

2 years ago

Doing research for insurance rates, it is found that those aged 30 to 49 drive an average of 38.7 miles per day with a standard

Alja [10]

Answer: 0.4302

Step-by-step explanation:

Given : Mean : $\mu=\text{38.7 miles }$

Standard deviation : $\sigma=\text{6.7 miles }$

Sample size : $n=60$

Also, these distances are normally distributed.

Then , the formula to calculate the z-score is given by :-

$z=\dfrac{x-\mu}{\sigma}$

For x=32.5

$\\\\ z=\dfrac{32.5-38.7}{6.7}=-0.925373134\approx-0.93$

For x=40.5

$\\\\ z=\dfrac{40.5-38.7}{6.7}=0.268656\approx0.27$

The p-value = $P(-0.93$

$=P(0.27)-P(-0.93)=0.6064198- 0.1761855=0.4302343\approx0.4302$

Hence, the required probability :-0.4302

3 0

3 years ago

One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla

Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0

3 years ago

to prepare a pesticide spray 3.5 lb of spray is added to 30 gal of Water how much pesticides spray should be added to spray tank

amm1812

I got 41.4 lbs if you round to the nearest tenths

6 0

3 years ago

URGENT

kupik [55]

The best way to approach this problem is to look at the graph of the given function. Replace values of x from 1 to 24 to indicate the numbers of hours in a day. As seen on the graph, there is only one point where the port is at high tide. That would be at 1:00 am.

Looking at the graph, it would be safe for the boats to be in the port when the graph levels off at around 10 to 24. That's from 10 am to before 12 midnight. Then, they would have to stay away between 12 midnight to before 10 am.

6 0

3 years ago