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3 years ago

10

Can i get help with this?

2 answers:

diamong [38]3 years ago

5 0

The answer is -6 and 8

nekit [7.7K]3 years ago

4 0

one point should go on 8 and another on -6

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Compute 1 + 2 + 3 +....+ 1,997 + 1,998 + 1,999

sp2606 [1]

Answer:

1 999 000

Step-by-step explanation:

Formula:

$1+2+3+.\ .\ .+n=\frac{n\times \left( n+1\right) }{2}$

………………………………………

Then

$1+2+3+....+1997+1998+1999=\frac{1999\times \left( 1999+1\right) }{2}$

$1+2+3+....+1997+1998+1999=\frac{1999\times \left( 2000\right) }{2}$

$1+2+3+....+1997+1998+1999=\frac{3998000 }{2}$

$1+2+3+....+1997+1998+1999=1999000$

7 0

2 years ago

Write the first five terms of the arithmetic sequence. a4 = 22, a10 = 64

dedylja [7]

Hello :
the general term is :
an = a1+(n-1)×d
or : an = ap +(n-p)×d......d is common diffrence
let : n=10 and p= 4
a10 = a4 +(10-4)×d
64 = 22 + 6d
6d = 42
d= 7
conclusion :
a4 = 22
a3 = 22-7=15
a2=15-7=8
a1= 8-7=1
<span>the first five terms : 1 , 8 , 15 , 22</span>

4 0

3 years ago

Lots of points plzzz help. What are m

lara31 [8.8K]

Answer:

???

Step-by-step explanation:

6 0

2 years ago

The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p

VLD [36.1K]

Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

$\chi^2_{1-\frac{\alpha}{2}}= 51.770$

Chi square critical value for upper tail =

$\chi^2_{\frac{\alpha}{2}}= 81.085$

80% confidence interval:

$\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13$

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0

3 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago