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3 years ago

Or is it c. 10.96 or d 13.0 (sorry the rest couldn’t fit in the picture) please help me I don’t get it.

Mathematics

1 answer:

solong [7]3 years ago

6 0

We have a right triangle with legs of 6 and 8. The hypotenuse is unknown so let it be x for now

Plug these three values
a = 6
b = 8
c = x
into the Pythagorean Theorem below. Then solve for x

a^2 + b^2 = c^2
6^2 + 8^2 = x^2
36+64 = x^2
100 = x^2
x^2 = 100
sqrt(x^2) = sqrt(100) ... apply the square root to both sides
x = 10

The final answer is 10 ft. This option isn't listed which makes me think your teacher made a typo somewhere.

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0.62 cents per pound

Step-by-step explanation:

2.48/4

to find price per pound which is the unit price

2.48/4=0.62

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3 years ago

How do you do this please help ?

Vladimir79 [104]

Answer:

Area: A = 42x - 49

Step-by-step explanation:

Area of rectangle = L * W

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4 years ago

Ayudaaaaa porfavor Convertir las unidades de medida 8 000 m = _______ km 4 000 g = _______ kg 3 000 m = ________ km 7 m = ______

jek_recluse [69]

Answer:

a) $x = 8\,km$ , b) $x = 4\,kg$ , c) $x = 3\,km$ , d) $x = 700\,cm$ , e) $x = 3\,cm$ , f) $x = 100\,mm$ , g) $x = 7\,mm$ , h) $x = 10000\,mL$ , i) $x = 8\,m$ , j) $x = 6000\,g$

Step-by-step explanation:

a) 1 kilómetro equivale a 1000 metros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 8000\,m \times \frac{1\,km}{1000\,m}$

$x = 8\,km$

b) 1 kilogramo equivale a 1000 gramos, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 4000\,g \times \frac{1\,kg}{1000\,g}$

$x = 4\,kg$

c) 1 kilómetro equivale a 1000 metros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 3000\,m \times \frac{1\,km}{1000\,m}$

$x = 3\,km$

d) 1 metro equivale a 100 centímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 7\,m \times \frac{100\,cm}{1\,m}$

$x = 700\,cm$

e) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 30\,mm \times \frac{1\,cm}{10\,mm}$

$x = 3\,cm$

f) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 10\,cm \times \frac{10\,mm}{1\,cm}$

$x = 100\,mm$

g) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 70\,mm\times \frac{1\,cm}{10\,mm}$

$x = 7\,mm$

h) 1 litro equivale a 1000 mililitros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 10\,L\times \frac{1000\,mL}{1\,L}$

$x = 10000\,mL$

i) 1 metro equivale a 100 centímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 800\,cm \times \frac{1\,m}{100\,cm}$

$x = 8\,m$

j) 1 kilogramo equivale a 1000 gramos, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 6\,kg\times \frac{1000\,g}{1\,kg}$

$x = 6000\,g$

8 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .