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3 years ago

Which expression represents 7 more than a number

Mathematics

2 answers:

marishachu [46]3 years ago

4 0

Answer:

2x plus 7

Step-by-step explanation:

Lera25 [3.4K]3 years ago

3 0

Answer n+7

Step-by-step explanation:

seven more than a number

so lets say we have a number n, how do we get seven more than this number?

we need to add seven.

this can be written as

n+7

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6+[11-{10+2-(8 of 2+1-3×4) }]

Vesna [10]

Answer:

Step-by-step explanation:

Step-by-step explanation:

6+[11-{10+2-(8×2+1-3×4)}]

6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)}

6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)} ]6+[11-[10+2-12}]

6+[11-{10+2-(8×2+1-3×4)}]6+[11-{10+2-(12)} ]6+[11-{10+2-12}]6+[11-{0}]

6+[11-0]

6+[11]

6+11

17 Ans.

7 0

3 years ago

In a book the characters are trying to determine whether the Poisson probabilistic model can be used to describe the locations t

oksian1 [2.3K]

Answer:

The estimated mean number of rockets hits in the region is 533.

Step-by-step explanation:

We are given the following information,

Number of rocket hits | Observed number of regions

0 | 228

1 | 214

2 | 94

3 | 32

4 | 7

5 | 0

6 | 0

7 | 1

We are asked to estimate the mean number of rocket hits in the region.

The mean or expected value is given by

$\mu = \sum (x \cdot P(x)) \\\\\mu = (0 \cdot 228) + (1 \cdot 214) + (2 \cdot 94) + (3 \cdot 32) + (4 \cdot 7) + (5 \cdot 0) + (6 \cdot 0) + (7 \cdot 1) \\\\\mu = 0 + 214 +188+ 96 +28 +0+ 0 + 7 \\\\\mu = 533$

Therefore, the mean number of rockets hits in the region is 533.

5 0

4 years ago

What is .7 of a mile

bearhunter [10]

Answer:

1232 yards or 3696 feet or 44352 inches or 0.7 of a mile

Step-by-step explanation:

4 0

3 years ago

A man's age is three years less than 4 times the age of his oldest daughter. What is age of his daughter

miskamm [114]

A=4×-3 I think this is the equation

4 0

4 years ago

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers