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3 years ago

What is 6.974 rounded to the nearest tenth place?

Mathematics

2 answers:

loris [4]3 years ago

7 0

Answer:

Step-by-step explanation:

The tenth place is the 9 in 6.974.

You round up because if the place next to it(the 7) is greater than 5 then you round up. If it is less than 5 the 7 becomes a 0.

But this is a special case because 6.9 would not become a 6.10 it would become a 7.

dezoksy [38]3 years ago

3 0

Answer:

Step-by-step explanation:

The digits six is in the 10th Pl. even though the other times place so you have to like go to the nine’s and if the ninth grade on the six Danny do you have to round up because the room was fine and hire you round up for or lower your round down to round up to a six stays at 7 because it doesn’t have to make any changes but if you want to go

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If the quotient is 12 and the dividend is 288 what is the divisor

WARRIOR [948]

To solve the question we proceed as follows;
dividend/divisor=quotient
from above we can say that:
divisor=dividend/ quotient
=288/12
=24
The divisor is 24

4 0

3 years ago

Complete the square for 5x^2 – 30x = 5.

Yuri [45]

First, subtract 5 from both sides, leaving you with 5 $x^{2}$ - 30x - 5 = 0
If you use the quadratic formula (a=5), (b=-30), (c=-5)

x=-b +/- √b²-4ac / 2a

x= -(30) +/- √(30)² - 4(5) * (-5) / 2(5)

x= 30 +/- √1000 / 10

x = 3 +/- √10

7 0

3 years ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the normal distribution and the central limit theorem, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

22% of couples meet online, hence p = 0.22.

A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

$\mu = p = 0.22$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338$

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is one subtracted by the p-value of Z when X = 0.25, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.25 - 0.22}{0.0338}$

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15, hence:

X = 0.2:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.2 - 0.22}{0.0338}$

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.15 - 0.22}{0.0338}$

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the normal distribution and the central limit theorem, you can check brainly.com/question/24663213

4 0

2 years ago

Solve for x. JKL 2x-4 8 16 7 56

mariarad [96]

you have to use intercept theorem, also known as Thales's theorem to slove this problem

7 0

1 year ago

Line BD is tangent to circle O at C, Arch AEC = 299, and ACE = 93. Find Angle DCE.

VladimirAG [237]

In triangle ACE,
we know C=93,E can be calulated by using arch angle AEC...what ever that is....,using this we get A=180-(E+93)

So, by alternate segment theorem, DCE= A.

thats all i can say.

7 0

2 years ago