I need a little help here !

Which fraction is equivalent to 1/4? A.6/18 b.12/60 c.25/100 d.20/100

fredd [130]

Answer:

c. 25/100

Step-by-step explanation:

The given fraction is 1/4.

Here we need to find the equivalent to 1/4.

It c. 25/100

Let me explain.

Multiply 1/4 by 25 both the numerator and the denominator.

1/4 *25/25 = 1*25 / 4*25

= 25/100

All other fractions are not equivalent to 1/4

Answer: c. 25/100

Thank you.

6 0

3 years ago

Read 2 more answers

Choose the symbol that makes a true statement. -3.8 _____ 0.1 > =

melomori [17]

Answer:0.1 is more than -3.8 so it would be >

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

17+36=z-37 what is z

sp2606 [1]

Answer:

The correct answer is 90

Step-by-step explanation:

17 + 36= z- 37 ADD 17 AND 36

53= z-37 BRING 37 OVER TO 53 AND ADD

z = 90 FINAL ANSWER

8 0

2 years ago

F(x)=-x+9 That’s it I just need someone to help me with my math homework

Vilka [71]

X intercept is (9,0)
Y intercept is (0,9)
Pls mark me as brainliest

7 0

2 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

2 years ago