What is 6.85 times 2 times 10?

Hi can someone help me with number 1 please!!

finlep [7]

Jenna is correct there are multiple numbers you can multiply to get rid of the fraction Not only 1 because there are more than 1 common denominators.

It even states it.

4 0

3 years ago

3+(2+8)2ndpower ÷4×(1/2) 4th power

PIT_PIT [208]

Answer:

412

Step-by-step explanation:

You have to follow PEMDAS and solve the problem carefully.

$\frac{3+(2+8)^{2} }{4*(\frac{1}{2} )^{2} } =\frac{3+(10)^{2} }{4*(\frac{1}{2} )^{2} } = \frac{3+100}{4*(\frac{1}{2} )^{2} } = \frac{103 }{4*(\frac{1}{2} )^{2} }$

$\frac{103 }{4*(\frac{1}{2} )^{2} } = \frac{103 }{4*(\frac{1}{16})} } = \frac{103 }{\frac{1}{4} } =\\103*4\\=412$

8 0

3 years ago

The teacher collected some snowfall data during one of the snow days last year. Estimate the total snowfall throughout the day.

Sedaia [141]

The total snowfall throughout the day is the expected value of the distribution.

The estimated total snowfall during the day is 30 inches

Start by calculating the average rate of snow collected on that day.

$Rate = \frac{0 + 1.5 + 2.5 + 2.0 +1.0 +0.5}{6}$

This gives

$Rate = \frac{7.5}{6}$

Divide 7.5 by 6

$Rate = 1.25$

There are 24 hours in a day.

So, the total snowfall is:

$Total = Rate \times Hours$

This gives

$Total = 1.25 \times 24$

Multiply

$Total = 30$

Hence, the estimated total snowfall during the day is 30 inches

Read more about expected values at:

brainly.com/question/13005177

3 0

2 years ago

A recent health report revealed that a woman with insurance spends an average of 2.3 days in the hospital following a routine ch

cupoosta [38]

Answer:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

$p_v =2*P(t_{30}>2.385)=0.0236$

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

$(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611$

$(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861$

Step-by-step explanation:

Data given and notation

$\bar X_{insurance}=2.3$ represent the mean for insurance

$\bar X_{No. ins}=1.9$ represent the mean withour insurance

$s_{Insurance}=0.6$ represent the sample standard deviation for the insurance case

$s_{No. Ins}=0.3$ represent the sample standard deviation for the No insurance case

$n_{Insurance}=16$ sample size for insurance

$n_{No. Iss}=16$ sample size for no insurance

t would represent the statistic (variable of interest)

$\alpha=0.01$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis: $\mu_{Insu}=\mu_{No. Ins}$

Alternative hypothesis: $\mu_{Ins} \neq \mu_{No. Ins}$

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

$t=\frac{\bar X_{Ins}-\bar X_{No.Ins}}{\sqrt{\frac{s^2_{Ins}}{n_{Ins}}+\frac{s^2_{No. Ins}}{n_{No. Ins}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385$

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

$df = n_{ins} +n_{No Ins}-2=16+16-2 =30$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{30}>2.385)=0.0236$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

$(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611$

$(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861$

6 0

3 years ago

50 PTS PLUS I WILL MARK BRAINLIEST

ch4aika [34]

Subtract 5x to both sides

7 0

3 years ago

Read 2 more answers