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3 years ago

What is the expanded form of .368?

Mathematics

2 answers:

JulsSmile [24]3 years ago

6 0

.368
= .3 + .06 + .008

schepotkina [342]3 years ago

5 0

.300 + .060 + .008
You would break each number's place value down and add them in a line, in decimals, you would do the exact same.

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110% of what number is 880?

Brums [2.3K]

I would make a proportion out of this to help solve it because I find proportions a bit easier to solve, so the proportion would be: 110/100=880/x

You would just solve it for 800

4 0

3 years ago

Solve this quadratic equation using the quadratic formula. 5 - 10x - 3x2 = 0

Olegator [25]

-3x^2 - 10x + 5 = 0

Quadratic formula:

-b +/- sqrt[b^2 - 4(a)(c)]/2(a)

So. . . 10 +/- sqrt[(10^2) - 4(-3)(5)]/2(-3)

10 +/- sqrt[100+60]/ -6

10 +/- sqrt[160]/-6

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3 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were born, a

Bas_tet [7]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

b)Yes, the proportion of girls is significantly different from 0.5.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

In the study 340 babies were born, and 289 of them were girls. This means that $n = 340, \pi = \frac{289}{340} = 0.85$

Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90$

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be effective?

b)Yes, the proportion of girls is significantly different from 0.5.

4 0

3 years ago

Please help! What's the area of this triangle? (Please put an answer that helps...)

LenKa [72]

Answer:

Step-by-step explanation:

see attachment below

3 0

3 years ago

Please help!! Question is in the picture!!

Debora [2.8K]

$\frac{17}{60}$ and $\frac{1}{4}$ are the experimental probabilities from the table.

Solution:

Total number of times spun the spinner = 60

Number of times getting 1 = 12

Number of times getting 2 = 17

Number of times getting 3 = 15

Number of times getting 4 = 16

Experimental probabilities from the table:

Probability of getting 1 = $\frac{12}{60}=\frac{1}{5}$

Probability of getting 2 = $\frac{17}{60}$

Probability of getting 3 = $\frac{15}{60}=\frac{1}{4}$

Probability of getting 4 = $\frac{16}{60}=\frac{4}{15}$

To determine which are the experimental probabilities from the table:

Option A: 15

This is not obtained above. So, it is not the experimental probability.

Option B: $\frac{17}{60}$

This is obtained in the probability of getting 2.

So, it is the experimental probability.

Option C: $\frac{1}{4}$

This is obtained in the probability of getting 3.

So, it is the experimental probability.

Option D: $\frac{7}{15}$

This is not obtained above. So, it is not the experimental probability.

Option E: $\frac{12}{43}$

This is not obtained above. So, it is not the experimental probability.

Hence $\frac{17}{60}$ and $\frac{1}{4}$ are the experimental probabilities from the table.

5 0

3 years ago