What is the volume of a solid that 2m, 2m, 9m, 9m, 6m?

Ksju [112]

Answer:

We conclude that the mean contents of cola bottles is more than or equal to the advertised 300 ml.

Step-by-step explanation:

We are given that Bottles of a popular cola drink are supposed to contain 300 ml of cola. The distribution of the contents is normal with standard deviation of 3 ml.

A student who suspects that the bottler is under-filling measures the contents of six bottles. The results are: 299.4, 297.7, 301.0, 298.9, 300.2, 297.0

Let $\mu$ = mean contents of cola bottles.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 300 ml {means that the mean contents of cola bottles is more than or equal to the advertised 300 ml}

Alternate Hypothesis, $H_A$ : $\mu$ < 300 ml {means that the mean contents of cola bottles is less than the advertised 300 ml}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

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

where, $\bar X$ = sample mean contents of cola bottle = $\frac{\sum X}{n}$ = 299.03 ml

$\sigma$ = population standard deviation = 3 ml

n = sample of bottles = 6

So, test statistics = $\frac{299.03-300}{\frac{3}{\sqrt{6} } }$

= -0.792

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -0.792 > -1.6449, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean contents of cola bottles is more than or equal to the advertised 300 ml.

8 0

3 years ago

I need help my question is -2 less than or equal to 3x-2 <10

Kisachek [45]

A n s w e r:

0 ≤ x < 4

4 0

3 years ago

A local high school has both male and female students.

Kitty [74]

Answer:

a. P(male) = 0.4

b. P(no sport and male) = 0.1

c. Unclear question (isn't it the same as b.?)

Step-by-step explanation:

The data below is what I've worked according to, which isn't very clear from the question so the answers are only correct if this is the correct table of data;

$\left[\begin{array}{ccc}&No \ Sports&Sports\\Female&10&32\\Male&7&21\end {array}\right]$

a.

$P(male) = \frac{7 + 21}{70} \\\\ = \frac{28}{70} \\\\ = \frac{2}{5}$

b.

Using the tree diagram in the picture;

$P(no \ sport \ and \ male) = \frac{2}{5} * \frac{1}{4} \\\\ = \frac{1}{10}$

8 0

2 years ago

What is the answer(FP!) 29/3 5/5 20/10 100/50

liubo4ka [24]

Answer:

123456789101112131415161718192021222324252627282930

Step-by-step explanation:

hahaha

8 0

2 years ago

You are on the south side of a river and you need to cross the river in order to get to a campsite. Not

Doss [256]

Answer:

Width of river = 0.25 mile

As length of river is not given , but it is given that campsite is located 1 mile east upstream, So length of river = 1 mile

Speed of mine in river while going upstream = 3 miles per hour

Speed of current as going Upstream = 1 miles per hour

Now Reduced speed of mine = Speed of mine in river while going upstream - Speed of current as going Upstream

= 3 - 1

= 2 miles per hour

1. There are two ways to reach campsite

(a) either walk 1 mile from starting point in east direction and then swim 0.25 mile in North direction to reach campsite.

Or

(b) From the starting point Considering a right triangle having, base=1 mile and altitude = 0.25 mile, then finding the length of hypotenuse using formula = $\sqrt{(altitude)^2 + (Base)^2$

= $\sqrt{.25^2+1^2}$

=√1.0625

=1.03078(approx)

⇒ you must walk a distance of 1 mile in the east direction and then swim 0.25 mile north direction to reach directly at campsite.[shown in the diagram]

7 0

3 years ago