4-0,9+(-0,11)-(-1,3) PLEASE HELP

Jessica saved a total of $58.50 to spend on school supplies. She wants to buy one backpack for $36.25 and she wants to spend the

Eduardwww [97]

Answer:

B. $36.25+$3.89x≤$58.50

Step-by-step explanation:

It has to be less than or equal to $58.50

4 0

3 years ago

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Please solve these asap. 50 points and will be marked as brainliest

Sergeu [11.5K]

Answer:

Hi myself Shrushtee.

Step-by-step explanation:

We are unable to see your answer clearly post it by scanning it ok post it again.

4 0

2 years ago

Please if you're feeling depressed or something, reach out to these people

Naddika [18.5K]

Answer:

thank you i will make sure to<3

3 0

2 years ago

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It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work i

Salsk061 [2.6K]

Answer:

$(-\infty, 0.2661)$

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

n=60 represent the sample size

X=12 represent the number of masks that had lenses pop out at 250°

$\hat p =\frac{12}{60}=0.2$ represent the estimated proportion of masks that had lenses pop out at 250°

p represent the population proportion of masks that had lenses pop out at 250°

Confidence =0.9 or 90%

$\alpha=0.1$ represent the significance level

Confidence interval

On this case we want a interval on this form : $(-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})$

So the critical value would be on this case $z_{\alpha}=1.28$ and we can use the following excel code to find it: "=NORM.INV(1-0.1,0,1)"

We found the lower limit like this:

$0.2 +1.28\sqrt{\frac{0.2 (1-0.2)}{60}}=0.2661$

And the interval would be: $(-\infyt, 0.2661)$

3 0

3 years ago

Suppose that in a senior college class of 500500 students, it is found that 179179 smoke, 228228 drink alcoholic beverages, 1

olga2289 [7]

Answer: a) 0.16, b) 0.058, and c) 0.856.

Step-by-step explanation:

Since we have given that

Number of students = 500

Number of students smoke = 179

Number of students drink alcohol = 228

Number of students eat between meals = 119

Number of students eat between meals and drink alcohol = 59

Number of students eat between meals and smoke = 72

Number of students engage in all three = 30

a) Probability that the student smokes but does not drink alcohol is given by

$P(S-A)=P(S)-P(S\cap A)\\\\P(S-A)=\dfrac{179}{500}-\dfrac{99}{500}\\\\P(S-A)=\dfrac{179-99}{500}\\\\P(S-A)=\dfrac{80}{500}\\\\P(S-A)=0.16$

b) eats between meals and drink alcohol but does not smoke.

$P((M\cap A)-S)=P(M\cap A)-P(M\cap S\cap A)\\\\P((M\cap A)-S)=\dfrac{59}{500}-\dfrac{30}{500}\\\\P((M\cap A)-S)=\dfrac{59-30}{500}\\\\P((M\cap A)-S)=\dfrac{29}{500}\\\\P((M\cap A)-S)=0.058$

c) neither smokes nor eats between meals.

$P(S'\cap M')=1-P(S\cup M)\\\\P(S'\cap M')=1-\dfrac{72}{500}\\\\P(S'\cap M')=\dfrac{500-72}{500}\\\\P(S'\cap M')=\dfrac{428}{500}=0.856$

Hence, a) 0.16, b) 0.058, and c) 0.856.

5 0

2 years ago