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Fittoniya [83]
3 years ago
7

What is the product of Three-fifths times StartFraction 2 Over 7 EndFraction?

Mathematics
2 answers:
Yuki888 [10]3 years ago
3 0

Answer:

<h3>StartFraction 6 Over 35 EndFraction</h3>

Step-by-step explanation:

Given the expression;

3/5 * 2/7

We are to find the product of bothh fractions

3/5 * 2/7

= (3*2)/(5*7)

= 6/35

Hence the correct result is 6/35 or StartFraction 6 Over 35 EndFraction

nevsk [136]3 years ago
3 0

Answer:

A

Step-by-step explanation:

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What is this math problem???<br><br> g(x)=1/20x(x-100)
Fudgin [204]

Answer:

the x intercept is (0,100) and the y intercept is (0,0)

Step-by-step explanation:

g(x)=1/20x(x-100)

g(0)=1/20(0)(0-100)

g(0)=1/20*0(0-100)

g(0)=100

6 0
3 years ago
25y + 150x = 250<br><br> find the Variables
fomenos

Answer: Step-by-step explanation:

25y + 150x = 250 (÷25)

y+6x=10

y=10-6x

example x=1 y=4

8 0
3 years ago
Let f ( x ) = 2 x − 1 , g ( x ) = 3 x , and h ( x ) = x ^2 + 1 , what is h( h ( 5) ) ?
strojnjashka [21]
First you would solve for h(5) by plugging in 5 as your x, then solving it.

h(5) = 5^2 + 1
h(5) = 25 + 1
h(5) = 26

Next you would multiply the 26 by the individual h, which is basically h(1).

h(1) = 1^2 + 1
h(1) = 2

Lastly you multiply your h(1) value by the h(5) value to get your answer.

h(1) • h(5) = 26 • 2
h[h(5)] = 52
7 0
4 years ago
Please help me ASAP!!!
vovangra [49]
Answer is D the last one

if the equation is y= 3 lx + 2l

if you solve for x then the zero is x= -2

if y intercept .. so x = 0 

then point is (0,6)

hope that helps
4 0
3 years ago
Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal distribution. A random sample
GuDViN [60]

Answer: (628.48,\ 661.52)

Step-by-step explanation:

Given : Sample size : n=16 , which is a small sample (, 30) so we use t-test.

Sample mean : \overline{x}=645 \text{ hours}

Standard deviation : \sigma = 31 \text{ hours}

Significance level : \alpha=1-0.95=0.05

Critical value : t_{n-1,\alpha/2}=2.131

The confidence interval for population mean is given by :-

\overline{x}\pm t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=645\pm(2.131)\dfrac{31}{\sqrt{16}}\\\\\approx645\pm16.52\\\\=(645-16.52,\ 645+16.52)\\\\=(628.48,\ 661.52)

Hence, a 95% confidence interval for the population mean \mu = (628.48,\ 661.52)

6 0
3 years ago
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