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3 years ago

PLEASEEEE HELPPP MEEE :(

Mathematics

1 answer:

stealth61 [152]3 years ago

4 0

Answer:

Question 5 is B

Because after I simplified I got B and its the only one that makes sense.

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Expression tha have the same vaule

Vedmedyk [2.9K]

I don't know if this is the answer that you are looking for, but would the answer be all solutions?

6 0

3 years ago

-1.2= y/2.4 -1.7<br> Solve for y

evablogger [386]

${\huge{\boxed{\sf{Question}}$

$-1.2= y\div2.4 -1.7$

Solve for y

${\huge{\boxed{\sf{Answer\:with\:explanation }}}$

let's put our equation first

$-1.2= y\div2.4 -1.7$

Flip the equation

$y\div2.4-1.7=-1.2$

Now follow the rule of $BOMDAS/PEMDAS$

${\huge{\boxed{\sf{BOMDAS }}}$

Brackets

Of multiplication

Division

Addition

Subtraction

${\huge{\boxed{\sf{PEMDAS}}}$

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

_________________________

${\huge{\boxed{\sf{Solve\:the\:equation}}}$

_________________________

$y\div2.4-1.7=-1.2$

${\underline{simplify}}$

$0.416667y-1.7=-1.2$

add $1.7$ to both sides

$0.416667y=0.5$

Divide both sides by $0.416667$

$y=1.2$

${\huge{\boxed{\sf{Answer}}}$

$y$ would equal $1.2$

4 0

3 years ago

PLEASEE HELP!! NUMBER 4!!

Illusion [34]

?????????????????????/ us a calculator to add up all the nub.

8 0

3 years ago

A satellite cable company charges an installation fee of $50 plus an additional $35.95 per month for service.

wolverine [178]

My answer would be 85.95

3 0

3 years ago

A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl

Marina86 [1]

Answer:

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 13

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.0123

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.0123\frac{279.29}{\sqrt{14}} = 224.85$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 664.14 - 224.85 = $439.29

The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

4 0

3 years ago