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3 years ago

Given △TSP, TS = SP = 10 cm TP = 12 cm Find: The area of △TSP

Mathematics

2 answers:

ivanzaharov [21]3 years ago

5 0

Answer:

22cm

Step-by-step explanation:

iogann1982 [59]3 years ago

4 0

Answer:

Step-by-step explanation:

if ts = 10 and sp = 10, use the formula 1/2 b x h therefore it will be 5 x 12=60, or 10 x 6 = 60. So the answer of the area of triangle tsp is 60.

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3 years ago

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natita [175]

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Step-by-step explanation:

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3 years ago

1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti

uysha [10]

Answer:

1) $0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

b) $ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=0.0826052$ represent the sample mean

$\mu$ population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom, given by:

$df=n-1=5-1=4$

The Confidence level is 0.95 or 95%, and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , the critical value would be using the t distribution with 4 degrees of freedom: $t_{\alpha/2}=2.776$