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alexandr1967 [171]
3 years ago
14

I need help please .

Mathematics
1 answer:
Simora [160]3 years ago
8 0

Answer:

the answer is 1 because kyut ko hahaha

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A cylinder with a height of 12 cm is inscribed in a sphere with a radius of 10 cm. Find the lateral area of the cylinder. Answer
zalisa [80]

Answer:

256π

Step-by-step explanation:

Given that:

The height of the cylinder: 12 cm

The radius of the sphere : 10

Let r is the radius of the cylinder, use Pytagon

10² = r² + 6²

<=> r²  = 10² - 6²  = 64

<=> r = 8

Hence,  the lateral area of the cylinder

L= 2πrh

= 2π8*16

= 256π

4 0
3 years ago
Line KL has an equation of a line y = 4x + 5. Which of the following could be an equation for a line that is perpendicular to li
djyliett [7]
To find a perpendicular slope (or line), the slope (in this case 4x) must be the opposite sign and its reciprocal, which is basically the fraction flipped upside down. Since 4 is technically 4/1, that fraction flipped is 1/4. And since you need to flip the sign too, instead of it being a positive number, it's negative. Your answer is -1/4x-8
3 0
3 years ago
Read 2 more answers
How does h
Agata [3.3K]

Answer:

h(t) = 16t changes 16 units every t. in the interval t=7 to t=2,

if its asking for:

\int\limits^2_7 {16t} \, dt

then it's

16(2) - 16(7) = 32 - 112 = -80

4 0
3 years ago
Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
Vesnalui [34]

Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

Since is a one left tailed test the p value would be:  

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0
3 years ago
Round 14.354 to the nearest one.
Gnoma [55]

Answer:

14

Step-by-step explanation: hope this helps :)

3 0
3 years ago
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