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lianna [129]
4 years ago
15

2100 +809 equals ?????

Mathematics
2 answers:
lord [1]4 years ago
6 0
2100 + 809 equals to 2909.
Naddika [18.5K]4 years ago
5 0
2100
+809
--------
 2909

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Find the sum: (2x^2 + x + 3) + (3x^2 + 2x + 1)
Montano1993 [528]

Answer:

5x^2 +  3x   +4

Step-by-step explanation:

(2x^2 + x + 3) + (3x^2 + 2x + 1)

Combine like terms

(2x^2 + 3x^2 + 2x+ x + 3 + 1)

5x^2 +  3x   +4

6 0
4 years ago
A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec
Ket [755]

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

(a)

The null hypothesis is:

<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

For 5% level of significance the two tailed critical value of <em>z</em> is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct the 95% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

3 0
3 years ago
Read 2 more answers
Please help me id really appreciate it
loris [4]

Answer:

Step-by-step explanation:

1. 2/5

2. 9/20

3. 0.43

4. 40.2%

5. 42%

6. 0.403

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

I truely hope this is right & i hope this helps you ! <3

5 0
3 years ago
The data in the table represents the average number of daylight hours each month in Springfield in 2015, rounded to the nearest
Lera25 [3.4K]
To do part B, use the equation that you wrote for part A.  We need to figure out what month number March of 2020 will be; we started in January 2015 at month 1; month 2 was Feb. 2015; month 3 was Mar. 2015; etc. to month 12 at Dec. 2015.  Then month 13 will be Jan. 2016; month 13+12=25 will be Jan. 2017; month 25+12=37 will be Jan. 2018; month 37+12=49 will be Jan. 2019; month 49+12=61 will be Jan. 2020; month 62 will be Feb. 2020; and month 63 will be March 2020.  

63 is the number you will substitute for x in the equation you've already written for part A.
8 0
3 years ago
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What are the original dimensions of the square piece of paper?
Inga [223]

Answer:

6root2 × 6root2

you have to use pythagoras

a²+b²=c²

c=12

12²=144

a²+b²=144

but it is a square so I'll just change a and b to x because it shows they are equal

x²+x²=144

2x²=144

x²=72

x=6root2

4 0
3 years ago
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