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4 years ago

What is 3/9 and 123/999 as a decimal

1 answer:

Nutka1998 [239]4 years ago

6 0

To convert fractions to decimals, divide the denominator into the numerator

3/9= .333333
123/999=.123123

hope this helps

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Which transformations could be performed to show that

emmainna [20.7K]

Answer:

no picture given

Step-by-step explanation:

no picture given, so it is impossible to say.

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3 years ago

Fill in the missing value to make the equations true.

Furkat [3]

Answer:

a) 7/4 b) 5 c) 2

Step-by-step explanation:

Logrithmic Rule for a and b

Let a, M, N be positive real numbers.

logaM - logaN = loga(M/N)

log9(7) - log9(4) = log9 (7/4)

logaM + logaN = logaMN

log2 (x) + log2(9) = log2(45)

x9=45

(x9)/9 = 45/9

x = 5

Change of base formula.

logb(x)=logd(b)/logd(x)

x log6(5) = log6(25) divide each term by log6(5)

x log6(5) / log6(5) = log6(25) / log6(5) Cancel common factor log6(5)

x = log6(25) / log6(5)

x = log6(5^2) / log6(5)

Expand log6(5^2) by moving 2 outside the logarithm.

x = 2log6(5) / log6(5) cancel the like term log6(5)

x = 2

7 0

4 years ago

Is it one solution or many solutions y=2x+7 and y=3x-1

vodka [1.7K]

Answer:

the answer is one solution

8 0

3 years ago

Read 2 more answers

Kamala has $90,000 in a savings

timama [110]

She will have 91,800

4 0

3 years ago

There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar am

Bas_tet [7]

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

$n = 479, \pi = \frac{130}{479} = 0.2714$

Hence the bounds of the interval are found as follows:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112$

The 95% confidence interval is (0.2316, 0.3112).

More can be learned about the z-distribution at brainly.com/question/25890103

7 0

2 years ago