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2 years ago

Now solve in simplest terms: 4/5+4/5 please help no links or files

Mathematics

1 answer:

xxTIMURxx [149]2 years ago

8 0

Answer:

1 and 3/5

Step-by-step explanation:

4/5+4/5=8/5

8÷5

=1 3/5

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If x=7 and y=−3, evaluate the following expression:3x−y/2

kogti [31]

Answer:

45/2

Step-by-step explanation:

3x-y/2

3(7)-(-3)/2

21+3/2

42/2+3/2

45/2

7 0

3 years ago

line J passes through points (-6,1) and (-3,6) . Line m is parallel to line j and passes through point (15,-1).what is the equat

VMariaS [17]

Slope of the line J
(6-1) / (-3-(-6))
5 / 3

so
m = 5/3

If this line passes through (15,-1)
y = mx + c
-1 = (5/3)*15 + c
-1 = 25 + c
c = -26

equation is
y = (5/3)x - 26

7 0

4 years ago

Guys Urgent Help me this example3.9 plzzz

patriot [66]

Answer:

A Hope this helps!!

Step-by-step explanation:

A because it doesn't matter who gets chosen for the committee there is still going to be more girls than boys but it doesn't have to be a must and it's not B because it would be 3 girls and 2 boys

6 0

3 years ago

You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%

bezimeni [28]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}$

$\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}$

6 0

4 years ago

Last year, the mean running time for a certain type of light bulb was 8.5 hours. This year, the manufacturer has introduced a ch

Free_Kalibri [48]

Answer:

H0 : μ = 8.5

H1 : μ > 8.5

1 sample t test ;

Test statistic = 2.53

Pvalue = 0.994

No, we fail to reject the Null ;

There is no significant evidence to support the Claim that the mean running time of light bulb is greater Tha last year.

Step-by-step explanation:

H0 : μ = 8.5

H1 : μ > 8.5

Test statistic :

(xbar - μ) ÷ (s/sqrt(n))

(8.7 - 8.5) ÷ (0.5 / sqrt40)

Test statistic = 2.53

The Pvalue :

P(Z < 2.53) = 0.9943

Pvalue = 0.994

Decison region :

Reject H0 ; if Pvalue < α

0.9943 > 0.05 ; We fail to reject the Null

5 0

3 years ago