Write each improper fraction as a mixed number 7/6 5/2 and 3/2

the sum of two numbers is 70. the larger number is five less than twice the smaller number . what are the two numbers

earnstyle [38]

X+y=70
x=2y-5
x=larger number
y=smaller number
now plug in the value of x to only have one variable in the equation...
Way to solve for y:
(2y-5)+y=70
3y-5=70
add 5 on both sides...
3y=75
divide by 3 on both sides
y=25
1st way to solve for x:
now plug in y in the original equation...
x+25=70
subtract 25 on both sides...
x=45
2nd way to solve for x:
plug in y for the second equation...
x=25(2)-5
x=50-5
x=45
answer: the two numbers are 45 and 25.

3 0

3 years ago

Giving brainlist to whoever answers

stealth61 [152]

Answer:

62.4 cm³

Step-by-step explanation:

Hope this helps :)

7 0

3 years ago

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro

-Dominant- [34]

Answer:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

Step-by-step explanation:

Data given and notation

$\bar X=1.6$ represent the sample mean

$s=0.46$ represent the sample deviation

$n=32$ sample size

$\mu_o =1.35$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis: $\mu =1.35$

Alternative hypothesis: $\mu \neq 1.35$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

P-value

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

5 0

3 years ago

{3x+y=7{-2x+y=-3 please

fredd [130]

Answer:

x= $\frac{4}{5}$