All categories

3 years ago

Please show how the following equasion Square root of 64+6/-2*-2 I cannot arrive at the answer of 9.5

Mathematics

2 answers:

Delicious77 [7]3 years ago

7 0

Answer:

$9.5$

Step-by-step explanation:

$\sqrt{64}+\frac{6}{-2\left(-2\right)}$

$\sqrt{64}+\frac{6}{2 \times 2}$

$8+\frac{6}{4}$

$\frac{19}{2}$

$=9.5$

frosja888 [35]3 years ago

5 0

Answer:

Hello!

I hope that this is the answer you are looking for

=8.09320

That is the rounded answer.

I hope that helped you!

Step-by-step explanation:

You might be interested in

What is the solution to the system graphed below?

lions [1.4K]

Answer:

I'm pretty sure it is (2,0)

3 0

2 years ago

A spider ate 25% more bugs this month then last month the spider ate 8 bugs last month

Ierofanga [76]

In this month spider ate 10 bugs.

Step-by-step explanation:

Given,

The spider ate 25% more bugs than last month.

It ate 8 bugs in last month.

To find the number of bugs it ate this month.

Solution

It ate more bugs = 8×25%

=8× $\frac{25}{100}$ = 2

In this month it ate = 8+2 = 10 bugs.

7 0

3 years ago

Read 2 more answers

Analyzing the Discrim

ANEK [815]

The values of x that will give the quadratic equation no real number solutions are -5 and 9

<h3>What are real roots of a quadratic equation?</h3>

Real roots of a quadratic equation are values of x, whose numbers can be represented on a number line.

Analysis:

We have - $x^{2}$ +3x +c = 0

$b^{2}$ - 4ac is called the discriminant of a quadratic equation.

if $b^{2}$ - 4ac < 0, then the quadratic equation has no real root.

where a = -1, b = 3 c = c

$(3)^{2}$ -4(-1)c < 0

9 + 4c < 0

4c < -9

c < -9/2

Which means values lower than -9/2 would give the solution no real root.

From, the options, only -5 would give no real solution, since it is lower than -9/2.

Learn more about real roots of a quadratic equation: brainly.com/question/7784687

#SPJ1

3 0

2 years ago

F(x) = (128/127)(1/2)x, x = 1,2,3,...7. determine the requested values: round your answers to three decimal places (e.g. 98.765)

Marrrta [24]

A.
$\mathbb P(X\le 1)=\mathbb P(X=1)=\dfrac{128}{127}\left(\dfrac12\right)^1=\dfrac{64}{127}$

b.
$\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\dfrac{64}{127}=\dfrac{63}{127}$

c.
$\mathbb E(X)=\displaystyle\sum_{x=1}^7 x\,f_X(x)=\frac{64}{127}\sum_{x=1}^7 x\left(\frac12\right)^{x-1}$

Suppose $f(y)=\displaystyle\sum_{x=0}^7 y^x$ . Then $f'(y)=\displaystyle\sum_{x=1}^7 xy^{x-1}$ . So if we can find a closed form for $f(y)$ , in terms of $y$ , we can find $\mathbb E(X)$ by evaluating the derivative of $f(y)$ at $y=\dfrac12$ .

$f(y)=\displaystyle\sum_{x=0}^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7$
$y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8$
$f(y)-y\,f(y)=y^0-y^8$
$(1-y)f(y)=1-y^8$
$f(y)=\dfrac{1-y^8}{1-y}$
$\implies f'(y)=\dfrac{7y^8-8y^7+1}{(1-y)^2}$
$\implies\mathbb E(X)=\dfrac{64}{127}f'\left(\dfrac12\right)=\dfrac{64}{127}\times\dfrac{247}{64}=\dfrac{247}{127}$

d.
$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

We find $\mathbb E(X^2)$ in a similar manner as in (c).

$\mathbb E(X^2)=\displaystyle\sum_{x=1}^7 x^2\,f_X(x)=\frac{32}{127}\sum_{x=1}^7x^2\left(\frac12\right)^{x-2}$

Now,

$f(y)=\displaystyle\sum_{x=0}^7y^x$
$\implies f'(y)=\displaystyle\sum_{x=1}^7xy^{x-1}$
$\implies f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$

We know that

$f''(y)=-\dfrac{42y^8-96y^7+56y^6-2}{(1-y)^3}$
$\implies f''\left(\dfrac12\right)=\dfrac{219}{16}$

We also have

$f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$
$f''(y)=\displaystyle\sum_{x=2}^7x^2y^{x-2}-\sum_{x=2}^7xy^{x-2}$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=2}^7x^2y^x-\sum_{x=2}^7xy^x\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\bigg(\sum_{x=1}^7x^2y^x-y\bigg)-\bigg(\sum_{x=1}^7xy^x-y\bigg)\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=1}^7x^2y^x-\sum_{x=1}^7xy^x\right)$

so that when $y=\dfrac12$ , we get

$\dfrac{219}{16}=4\left(\dfrac{127}{128}\mathbb E(X^2)-\dfrac{127}{128}\mathbb E(X)\right)\implies\mathbb E(X^2)=\dfrac{685}{127}$

Then

$\mathbb V(X)=\dfrac{685}{127}-\left(\dfrac{247}{127}\right)^2=\dfrac{25,986}{16,129}$

6 0

3 years ago

Please help me with this math equation 6d - 11/2 = 2d - 13/2

crimeas [40]

Step 1: Simplify both sides of the equation.6d+−112=2d+−132Step 2: Subtract 2d from both sides.6d+−112−2d=2d+−132−2d4d+−112=−132Step 3: Add 11/2 to both sides.4d+−112+112=−132+1124d=−1Step 4: Divide both sides by 4.4d4=−14d=−14

4 0

3 years ago