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3 years ago

Suppose a fair die is rolled eight times. find numerical values for the expectation

Mathematics

1 answer:

hodyreva [135]3 years ago

3 0

Your possiblity is 1/6 then you multiply that by tries
so 1/6 ×8

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I need help with this question. I cant find the surface area of this square pyramid. Please help me.

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25 bc 5 times 5 is 25

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2 years ago

Lola takes the train from Paris to Nice. The distance between the two cities is about 921,000 meters. If the train travels at a

Andru [333]

Distance =rate/time so your answer is 3 hours first you have to convert meters to kilometers 921000 meters = 921 Km so you have like terms to calculate so your distance becomes 921 Km then divide by the speed 907÷3 = 3 so your answer is 3 hrs

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2 years ago

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a bag contains a-r-I-t-h-m-e-t-I-c. Amelia chooses a tile without looking and doesn’t replace it. She chooses a second tile with

olga nikolaevna [1]

<span>The first time she chooses, her chances are 2/10 that she will get an I tile. With not replacing the tile, her second chance will be 1/9. Multiply the two fractions together gives 2/90 = 1/45
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2 years ago

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

3 years ago

Unit 7 polygons & quadrilaterals homework 3: rectangles Gina Wilson answer key

Irina-Kira [14]

Answer:

In the image attached you can find the Unit 7 homework.

We need to findt he missing measures of each figure.

Notice that the first figure is a rectangle, which means opposite sides are congruent so,

VY = 19

WX = 19

YX = 31

VW = 31

To find the diagonals we need to use Pythagorean's Theorem, where the diagonals are hypothenuses.

$VX^{2}=19^{2}+31^{2}\\ VX=\sqrt{361+961}=\sqrt{1322} \\VX \approx 36.36$

Also, $YW \approx 36.36$ , beacuse rectangles have congruent diagonals, which intercect equally.

That means, $ZX = \frac{VX}{2} \approx \frac{36.36}{2}\approx 18.18$

Figure number two is also a rectangle.

If GH = 14, that means diagonal GE = 28, because diagonals intersect in equal parts.

Now, GF = 11, because rectangles have opposite sides congruent.

DF = 28, because in a reactangle, diagonals are congruent.

HF = 14, because its half of a diagonal.

To find side DG, we need to use Pythagorean's Theorem, where GE is hypothenuse

$GE ^{2}=11^{2}+DG^{2}\\28^{2}-11^{2}=DG^{2}\\DG=\sqrt{784-121}=\sqrt{663}\\ DG \approx 25.75$

This figure is also a rectangle, which means all four interior angles are right, that is, equal to 90°, which means angle 11 and the 59° angle are complementary, so

$\angle 11 +59\°=90\°\\\angle 11=90\°-59\°\\\angle 11=31\°$

Now, angles 11 and 4 are alternate interior angles which are congruent, because a rectangle has opposite congruent and parallel sides.

$\angle 4 = 31\°$

Which means $\angle 3 = 59\°$ , beacuse it's the complement for angle 4.

Now, $\angle 6 = 59\°$ , because it's a base angle of a isosceles triangle. Remember that in a rectangle, diagonals are congruent, and they intersect equally, which creates isosceles triangles.