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2 years ago

What is 6 1/4 written as a decimal? Please help me I am very confused.

Mathematics

2 answers:

MrRa [10]2 years ago

7 0

Answer: 6.2

Step-by-step explanation:

charle [14.2K]2 years ago

4 0

Answer:

6.25

Step-by-step explanation:

1. Multiply the whole number by the denominator

2. Add the product you got to the numerator

3. Divide the sum by the denominator

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What is eight g plus ten equals thirty five plus three g

gtnhenbr [62]

8g+10=35+3g You need to find what number g is, just plug in random numbers till both sides are equal

8 0

3 years ago

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that $t_{\alpha/2}=$

Now we have everything in order to replace into formula (1):

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

5 0

3 years ago

Several years ago, 39% of parents who had children in grades K-12 were satisfied with the quality of education the students r

BaLLatris [955]

<h2>Answer with explanation:</h2>

Let p be the population proportion of parents who had children in grades K-12 were satisfied with the quality of education the students receive.

Given : Several years ago, 39% of parents who had children in grades K-12 were satisfied with the quality of education the students receive.

Set hypothesis to test :

$H_0: p=0.39\\\\H_a :p\neq0.39$

Sample size : n= 1055

Sample proportion : $\hat{p}=\dfrac{466}{1055}=0.441706161137\approx0.44$

Critical value for 95% confidence : $z_{\alpha/2}=1.96$

Confidence interval : $\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

$0.44\pm (1.96)\sqrt{\dfrac{0.39(1-0.39)}{1055}}\\\\0.44\pm0.0299536805135\\\\0.44\pm0.03\\\\=(0.44-0.03, 0.44+0.03)\\\\=(0.41,\ 0.47)$

Since , Confidence interval does not contain 0.39.

It means we reject the null hypothesis.

We conclude that 95% confidence interval represents evidence that parents' attitudes toward the quality of education have changed.

8 0

3 years ago

Simplify completely the quantity 3 times x to the 4th power plus 5 times x to the 2nd power plus 2 times x all over x. (2 points

BartSMP [9]

3x^3+5x+2 is the answer

5 0

3 years ago

Read 2 more answers

An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From pre

melamori03 [73]

Answer:

We conclude that a compact microwave oven consumes a mean of more than 250 W.

Step-by-step explanation:

We are given that an appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W with a population standard deviation of 15 W.

They take a sample of 20 microwave ovens and find that they consume a mean of 257.3 W.

Let $\mu$ = mean power consumption for microwave ovens.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 250 W {means that a compact microwave oven consumes a mean of no more than 250 W}

Alternate Hypothesis, $H_A$ : $\mu$ > 250 W {means that a compact microwave oven consumes a mean of more than 250 W}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

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

where, $\bar X$ = sample mean power consumption for ovens = 257.3 W

σ = population standard deviation = 15 W

n = sample of microwave ovens = 20

So, the test statistics = $\frac{257.3-250}{\frac{15}{\sqrt{20} } }$

= 2.176

The value of z test statistics is 2.176.

Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.

Since our test statistic is more than the critical value of t as 2.176 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that a compact microwave oven consumes a mean of more than 250 W.

7 0

2 years ago