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3 years ago

10 POINTS! EASY, ORDER OF OPERATIONS. PLEASE HURRY.

Mathematics

2 answers:

Fudgin [204]3 years ago

7 0

Additional information:

First U have to use the BODMAS rule.

B=Bracket

O=Of

D=division

M=Multiplication

A=Addition

S=Subtraction

If u use the BODMAS rule then it will be easy to solve....

Look at the attached picture ⤴

Hope it will help you....:)

<h3>✴✳Have a great day✳✴</h3>

WARRIOR [948]3 years ago

5 0

A) 1.2 +3.1×2 -(2.7 +0.6)/3
.. = 1.2 +3.1×2 -3.3/3
.. = 1.2 +6.2 -1.1
.. = 6.3

b) 9.9 +5×4.6
.. = 9.9 +23
.. = 32.9

c) 7×(6 +7.1)
.. = 7×13.1
.. = 91.7

Any good calculator can properly solve these.

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A foam toy is 2 inches wide. It doubles in size for every minute it is in water. Write an expression for the width of the toy af

Solnce55 [7]

it would be 2(2^5), which is two times two to the fifth power. you take the original size and double it for every minute, which is like squaring it, but to the power of five, because it doubles once every minute for five minutes. i hope this makes sense!:)

7 0

3 years ago

Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy

Alborosie

Answer:

H 0 : μ = 3.1 H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

$t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66$

$df = n-1=75-1 =74$

The p value is:

$p_v = P(t_{74} >8.66) =3.65x10^{-13}$

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1 H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

$\bar X = 3.13 , s =0.03 , n =75$

The statistic to check the hypothesis is given by:

$t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

Replacing the info given we got:

$t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66$

The degrees of freedom are given by:

$df = n-1=75-1 =74$

The p value since is a right tailed ted is given by:

$p_v = P(t_{74} >8.66) =3.65x10^{-13}$

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

6 0

3 years ago

For the equations given below, which statement is true?

allochka39001 [22]

Answer: C. The equations have the same solution because the second equation can be obtained by adding 6 to both sides of the first equation.

Step-by-step explanation:

You know that the first equation is:

$-3x-8=19$

And the second equation is:

$-3x-2=25$

According to the Addition property of equality:

If $a=b$ ; then $a+c=b+c$

Then, you can add 6 to both sides of the first equation to keep it balanced. Then, you get:

$-3x-8=19\\\\-3x-8+(6)=19+(6)$

$-3x-2=25$

Therefore, you can observe that the second equation can be obtained by adding 6 to both sides of the first equation, therefore, the equations have the same solution.

If you want to verify this, you can solve for "x" from both equations:

- First equation:

$-3x-8=19\\\\-3x=19+8\\\\x=\frac{27}{-3}\\\\x=-9$