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3 years ago

What is the equivalent for 4/6

Mathematics

2 answers:

Aleksandr [31]3 years ago

5 0

Answer:

$\frac{4}{6} \\ = \frac{2}{3} \\ = 0.66$

Kazeer [188]3 years ago

3 0

Answer:

2/3

Step-by-step explanation:

please mark this answer as brainliest

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What feature of the box and whisker plot does the IQR describe

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The IQR tells you how spread out the "middle" values are. It can also be used to figure out if there are any outliers.

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3 years ago

Order the decimal greatest to least 8.31,8.295, 8.06,8.3

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Answer:

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2 years ago

Read 2 more answers

5. Suppose that a particular candidate for public office is in fact favored by p = 48% of all registered voters. A polling organ

mart [117]

Answer:

Probability that the sample proportion will be greater than 0.5 is 0.8133.

Step-by-step explanation:

We are given that the a particular candidate for public office is in fact favored by p = 48% of all registered voters. A polling organization is about to take a simple random sample of voters and will use the sample proportion to estimate p.

Suppose that the polling organization takes a simple random sample of 500 voters.

Let $\hat p$ = sample proportion

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion = 48%

n = sample of voters = 500

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample proportion will be greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.50) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\frac{0.50-0.48}{\sqrt{\frac{0.50(1-0.50)}{500} } }$ ) = P(Z < 0.89) = 0.8133

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.89 in the z table which has an area of 0.8133.

Therefore, probability that the sample proportion will be greater than 0.50 is 0.8133.

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3 years ago

In a circle with a radius of 5cm and is centered at point O, angle AOB intercepts arc AB. Arc AB has a length of 10cm. What is t

Elenna [48]

circumference of circle = 2*pi*r given r = 5 so circumference = 31.42 now we know 360 degree is uniformly distributed on this circumference so x/360=10/31.42 so x= 114.65 so option C.) 114.65 is the answer

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2 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider