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3 years ago

Get this right and you get brainliest

Mathematics

1 answer:

blagie [28]3 years ago

5 0

Answer:

2 1/3 planes

Step-by-step explanation:

3.5 lego planes / 60 minutes

So fraction =

3.5/60

In 40 minutes he will build x planes

So the fraction equation is:

3.5/60 = x/40

Solve:

Cross multiply =

140 = 60x

x = 2 1/3

2 1/3 planes

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-Chetan K

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Drove 350 miles to her grandmother. The trip took her 5 1/4 hours. What was her average speed in miles per hours?

VikaD [51]

350mi/5.25h = 66.66 mph

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3 years ago

Write a polynomial in standard form that meets the following conditions. Assume a=1 and your function is f(x), The zeros are 5 a

jeka57 [31]

Polynomials are equations that uses variables and several terms

The polynomial in standard form is f(x) = x^2 - x -20

<h3>How to determine the polynomial</h3>

The polynomial has 2 zeros.

So, the form of the polynomial is:

f(x) = a(x - x1)(x - x2)

The zeros of the polynomial are 5 and -4.

So, the equation becomes

f(x) = a(x - 5)(x + 4)

The value of a = 1.

So, we have;

f(x) = 1(x - 5)(x + 4)

This gives

f(x) = (x - 5)(x + 4)

Expand

f(x) = x^2 - x -20

Hence, the polynomial in standard form is f(x) = x^2 - x -20

Read more about polynomials at:

brainly.com/question/2833285

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2 years ago

A tire manufacturer warranties its tires to last at least 20,000 miles orâ "you get a new set ofâ tires." In itsâ experience, a

Y_Kistochka [10]

Answer:

Probability that a set of tires wears out before 20,000 miles is 0.1151.

Step-by-step explanation:

We are given that a tire manufacturer warranties its tires to last at least 20,000 miles or "you get a new set of tires." In its past experience, a set of these tires last on average 26,000 miles with S.D. 5,000 miles. Assume that the wear is normally distributed.

<em>Let X = wearing of tires</em>

So, X ~ N( $\mu=26,000,\sigma^{2}=5,000^{2}$ )

Now, the z score probability distribution is given by;

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

where, $\mu$ = average lasting of tires = 26,000 miles

$\sigma$ = standard deviation = 5,000 miles

So, probability that a set of tires wears out before 20,000 miles is given by = P(X < 20,000 miles)

P(X < 20,000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20,000-26,000}{5,000}$ ) = P(Z < -1.2) = 1 - P(Z $\leq$ 1.2)

= 1 - 0.88493 = 0.1151

Therefore, probability that a set of tires wears out before 20,000 miles is 0.1151.

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3 years ago

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charle [14.2K]

Step-by-step explanation:

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