All categories

3 years ago

Missing pattern 24,21, blank 15 blank ,blank

2 answers:

8 0

To figure out this pattern, let's see if there's a constant rule that this sequence follows.
From 24 to 21, you're subtracting 3.
Let's try n-3, and fill in the blanks as we go.
24 - 3 = 21.
21 - 3 = 18.
18 - 3 = 15.
15 - 3 = 12.
12 - 3 = 9.
It seems that n-3 works as a rule, so your rule is:
N-3.
The missing blanks are:
18, 12, and 9.
I hope this helps!

Flura [38]3 years ago

5 0

The pattern would be n-3

You might be interested in

I will give brainly Help! Itss a Math Question !!! image is attached below

3241004551 [841]

Answer:

ratio of GEOM to TEYR is $\frac{6}{4}:1$

Therefore OE is equal to:

$8\cdot\frac{6}{4}$

$=12$

12 units is the length of OE, so the final answer is:

12.0 Units.

5 0

1 year ago

Read 2 more answers

Tşk şimdiden 17 puan

Kamila [148]

Answer:

um what-?

Step-by-step explanation:

5 0

2 years ago

I need to know this plz answer quick thanks - dely

Tpy6a [65]

120/12=x
120/12=10
Then answer to this problem is B

7 0

3 years ago

7 1/2 divided by 3/4

vagabundo [1.1K]

Answer:

7 1/2 divided by 3/4= 10

5 0

3 years ago

An environmentalist wants to find out the fraction of oil tankers that have spills each month. Step 2 of 2 : Suppose a sample of

rewona [7]

Answer:

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 333 tankers is drawn. Of these ships, 257 did not have spills.

333 - 257 = 76 have spills.

This means that $n = 333, \pi = \frac{76}{333} = 0.228$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 - 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.199$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 + 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.257$

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

6 0

2 years ago