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Roman55 [17]
3 years ago
8

PLEASE HELP The areas of two similar triangles are 50 dm2, and 32 dm2. The sum of their perimeters is 117 dm. What is the perime

ter of each of these triangles?
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

65 dm and 52 dm

Step-by-step explanation:

If the scale factor of the sides is k, then the scale factor of the areas is k^2.

The scale factor of the areas is (32 dm^2)/(50 dm^2) = 0.64 = k^2

The scale factor of the sides is k = \sqrt(0.64) = 0.8

The perimeters are in a ratio of 1:0.8

x + 0.8x = 117

1.8x = 117

x = 65

0.8x = 0.8(65) = 52

The perimeters are 65 dm and 52 dm.

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Que es lo mismo que 1u
Paul [167]

Answer:

Una unidad rack o simplemente U es una unidad de medida usada para describir la altura del equipamiento preparado para ser montado en un rack de 19 ó 23 pulgadas de ancho. Una unidad rack equivale a 1,75 pulgadas (44.45 mm) de alto.

Una unidad de rack se escribe normalmente como "1U"; del mismo modo dos unidades se escribe "2U" y así sucesivamente. La altura de una pieza del equipamiento de un rack es frecuentemente descrita como un número en "U".

Step-by-step explanation:

3 0
3 years ago
The breaking strengths of cables produced by a certain manufacturer have a mean, , of pounds, and a standard deviation of pounds
olchik [2.2K]

The question is incomplete. The complete question is :

The breaking strengths of cables produced by a certain manufacturer have a mean of 1900 pounds, and a standard deviation of 65 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 150 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1902 pounds. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the mean breaking strength has increased?

Solution :

Given data :

Mean, μ = 1900

Standard deviation, σ = 65

Sample size, n = 150

Sample mean, $\overline x$ = 1902

Level of significance = 0.01

The hypothesis are :

$H_0 : \mu = 1900$

$H_1 : \mu > 1900$

Test statics :

We use the z test as the sample size is large and we know the population standard deviation.

$z=\frac{\overline x - \mu}{\sigma / \sqrt{n}}$

$z=\frac{1902-1900}{65 / \sqrt{150}}$

$z=\frac{2}{5.30723}$

$z=0.38$

Finding the p-value:

P-value = P(Z > z)

             = P(Z > 0.38)

             = 1 - P(Z < 0.38)

From the z table. we get

P(Z < 0.38) = 0.6480

Therefore,

P-value = 1 - P(Z < 0.38)

            = 1 - 0.6480

             = 0.3520

Decision :

If the p value is less than 0.01, then we reject the H_0, otherwise we fail to reject  H_0.

Since the value of p = 0.3520 > 0.01, the level of significance, then we fail to reject  H_0.

Conclusion :

At a significance level of 0.01, we have no sufficient evidence to support that the mean breaking strength has increased.

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mamaluj [8]
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