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3 years ago

12

What is the volume of a cylinder, in cubic cm, with a height of 19cm and a base diameter of 6cm? Round to the nearest tenths pla

ce
Answer: V= 3 cm

1 answer:

SVETLANKA909090 [29]3 years ago

6 0

Answer:

It is about 537.2

Step-by-step explanation:

V= π · r^2 · h

V= π · 3^2 · 19

V≈ 537.2

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Adratic Formula

maw [93]

Answer:

Step-by-step explanation:

discriminant=5²-4×1×7=25-28=-3<0

no real number solution.

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3 years ago

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PLEASE ANSWER ASAP

blondinia [14]

Answer:

I've attached the Answer

Step-by-step explanation:

here it is

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3 years ago

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In the problem below, AB, CD, and EF are two-digit numbers, where A, B, C, D, E and F represent distinct digits from 1 to 9. Is

Oksi-84 [34.3K]

Answer:

<em>When both the conditions hold true, F is prime.</em>

Step-by-step explanation:

AB, CD, and EF are two-digit numbers, where A, B, C, D, E and F represent distinct digits from 1 to 9.

AB

+ CD

--------

EF

1st condition, B and D are consecutive.

Adding B and D gives us F.

Possible values can be (F being the unit value after adding not considering the carry over):

B + D = F

1+2=3

2+3=5

3+4=7

4+5=9

5+6=1

6+7=3

7+8=5

8+9=7

Here F is not prime (because 9 is not prime).

Now, let us consider the 2nd condition as well.

i.e. C = 8

For the following

AB

+ CD

--------

EF

C is 8 then A must be 1 because any value other than 1 for A will make the sum of A and C greater than 9 and there will be a carry which is not the case here.

So, E = 8 + 1 = 9

Now, B and D are consecutive and can not be 1, 8 or 9.

So, possible values are:

B + D = F

2 + 3 = 5

3 + 4 = 7

Here F is prime.

So, when both the conditions hold true, F is prime.

3 0

2 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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2 years ago

What is the difference? 4/3 - 7/9 a. - 1/2 b. -1/3 c. 5/9 d 11/12

ANTONII [103]

4/3-7/9

= (36-21) /27

= 15/27

= 5/9 which is C

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3 years ago

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