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3 years ago

14

in equivalent ratios , if the numerator of the first ratio is greater than the denominator of the first ratio , then the numerat

or of the second ratio is greater than the denominator of the second ratio

2 answers:

Murljashka [212]3 years ago

8 0

less than the denominator in the second ratio

REY [17]3 years ago

4 0

Answer:The denominator would be less then the second ratio.

Hope this helps!

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Find the slop of (10,5) (18,9)

Arada [10]

Answer:

1/2

Step-by-step explanation:

We can find the slope given two points by

m = (y2-y1)/(x2-x1)

= (9-5)/ (18-10)

= 4/8

= 1/2

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what differences can be found when contrasting the mood of third person acc with that of claudettes first person account?

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Answer: The mood of the third-person account is less emotional and more matter-of-fact. The mood of Claudette's account is less emotional and more matter-of-fact.

Step-by-step explanation:

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3 years ago

Find the value of x please and thank you

Pachacha [2.7K]

Given:

In triangle DEF, HG is parallel to DF.

To find:

The value of x.

Solution:

In triangles DEF and GEH,

$\angle DE F\cong \angle GEH$ (Common angle)

$\angle EDF\cong \angle EGH$ (Corresponding angle)

$\Delta DE F\sim \Delta GEH$ (By AA property of similarity)

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$\dfrac{DE}{GE}=\dfrac{DF}{GH}$

$\dfrac{9+(x+5)}{x+5}=\dfrac{12}{6}$

$\dfrac{x+14}{x+5}=2$

$x+14=2(x+5)$

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Isolating variable terms, we get

$x-2x=-14+10$

$-x=-4$

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Therefore, the value of x is equal to 4.

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2 years ago

Every Wednesday, Wayne works out for 3/4 of an hour at the fitness center. Every Saturday, he goes to the fitness center again a

matrenka [14]

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3 years ago

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Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally

rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = <u><em>thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.</em></u>

So, X ~ Normal( $\mu=5.0,\sigma^{2} =1.7^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean thickness = 5.0 mm

$\sigma$ = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

P(X < 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{3.0-5.0}{1.7}$ ) = P(Z < -1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

P(X > 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ > $\frac{7.0-5.0}{1.7}$ ) = P(Z > 1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X $\leq$ 3.0 mm)

P(X < 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{7.0-5.0}{1.7}$ ) = P(Z < 1.18) = 0.881

P(X $\leq$ 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{3.0-5.0}{1.7}$ ) = P(Z $\leq$ -1.18) = 1 - P(Z < 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

4 0

3 years ago