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3 years ago

9

The currency in Mexico is the Peso. The exchange rate approximately 11.1 Pesos for $3.40 At this rate, how many dollars would yo

u get if you exchanged 39.8 Pesos

1 answer:

Ymorist [56]3 years ago

7 0

You would get 12.19 dollars

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Eight balls numbered from 1 to 8 are placed in an urn. One ball is selected at random. Find the probability that it is

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Answer:

$\frac{7}{8}$

Step-by-step explanation:

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3 years ago

svet-max [94.6K]

Is this multiple choice

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3 years ago

HELP WILL GIVE BRAINLIST

Step2247 [10]

Answer:

Simplify {4}^{2}42 to 1616.

-16+2\times -4\times -5-2\times {2}^{3}−16+2×−4×−5−2×23

Simplify 2\times -42×−4 to -8−8.

-16-8\times -5-2\times {2}^{3}−16−8×−5−2×23

Simplify 8\times -58×−5 to -40−40.

-16-(-40)-2\times {2}^{3}−16−(−40)−2×23

Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}xaxb=xa+b.

-16-(-40)-{2}^{4}−16−(−40)−24

Simplify {2}^{4}24 to 1616.

-16-(-40)-16−16−(−40)−16

Remove parentheses.

-16+40-16−16+40−16

Simplify -16+40−16+40 to 2424.

24-1624−16

Simplify.

8

Step-by-step explanation:

hope it helps :)

6 0

3 years ago

Alaska has a land area of about 1,700,000 km2. florida has a land area 1/10 the size of alaska. what is the la nd area of florid

AlekseyPX

All you have to do is multiply 1,700,000 by .1 or you could divide 1,700,000 by 10 if you prefer that. The answer would be 170,000.

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4 years ago

A USA Today/CNN/Gallup survey of 369 working parents found 200 who said they spend too little time with their children because o

KengaRu [80]

Answer: a. 0.542

b. 0.051

c. $(0.491,\ 0.593)$

Step-by-step explanation:

Let p be the population proportion of parents spend too little time with their children because of work commitments

As per given , we have

n= 369

$\hat{p}=\dfrac{200}{369}=0.542005420054\approx0.542$

Since the sample proportion is the best estimate for the population proportion.

So , the point estimate of the proportion of the population of working parents who feel they spend too little time with their children because of work commitments = 0.542

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

Margin of error : $E=z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$E=(1.96)\sqrt{\dfrac{0.542(1-0.542)}{369}}\\\\=0.0508364564574\approx0.051$

The margin of error = 0.051

Confidence interval : $\hat{p}\pm E$

$0.542\pm0.051\\\\=(0.542-0.051,\ 0.542+0.051)\\\\=(0.491,\ 0.593)$

95% confidence interval estimate of the population proportion of working parents who feel they spend too little time with their children because of work commitments : $(0.491,\ 0.593)$

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3 years ago