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3 years ago

Aa right triangle has sides measuring 8 centimeteres, 15 centimeters and 17 centimeters. What is the area of the piece

Mathematics

1 answer:

stepladder [879]3 years ago

5 0

Answer:

Area = 127.5cm²

Step-by-step explanation:

Given that,

Length of first side = 8 cm

Length of second side = 15 cm

Length of third side = 17cm

To find,

Area of triangle

to find which of the sides is hypotenuse use Pythagorus theorem

hypotenuse² = height² + base²

17² = 8² + 15²

289 = 289

hence 17 is the hypotenuse and other two are height and base

<h2>Area of triangle = 1/2 height * base</h2>

Area of triangle = 1/2 15 * 17

Area of triangle = 1/2 (255)

Area of triangle = 127.5

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4 years ago

Read 2 more answers

Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11

Bond [772]

Answer:

a) Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

b) $t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

c) $12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

Step-by-step explanation:

Information given

$\bar X=12.50$ represent the mean for the daily iron intake

$s=4.75$ represent the sample deviation

$n=51$ sample size

$\mu_o =14.44$ represent the reference value

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Part a

We want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:

Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

Part b

Since we don't know the population deviation the statistic would be given by:

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

Replacing the info we got

$t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

Part c

The confidence interval would be given by:

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