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3 years ago

What is 22013.2220% + 12345.078965%+85$

2 answers:

6 0

<span>22013.2220 + 12345.078965%+85= </span>2739647.86064

You can convert the number to a percent or $ ur own self.
Ohh and I am pretty sure that NO math teacher maked u answer these types of problems!! :D

Mamont248 [21]3 years ago

3 0

220.132220 + 123.45078965 + 85 = 428.58300965

I turned the percentages into regular decimals by moving the decimal to the left 2 places and then I just added the values

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A car is traveling at a rate of 99 kilometers per hour. what is the car's rate in kilometers per minute? how many kilometers wil

gayaneshka [121]

there are 60 minutes per hour

99/60 = 1.65 km per minute

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4 years ago

A stockbroker has kept a daily record of the value of a particular stock over the years and finds that prices of the stock form

amid [387]

Answer:

$12.43

Step-by-step explanation:

Given :

Mean = $8.52

Standard deviation, = $2.38

Stock price which falls beyond 0.05 of the distribution is at the 95th percentile

The 95th percentile distribution has a Pvalue of 1.645 (standard normal table)

We obtain the value of x, with z = 1.645

Using the Zscore relation :

Zscore = (score - mean) / standard deviation

1.645 = (score - 8.52) / 2.38

Cross multiply :

1.645 * 2.38 = score - 8.52

3.9151 = score - 8.52

Score = 8.52 + 3.9151

Score = $12.4351

Stock price beyond 0.05 is $12.43

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3 years ago

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olya-2409 [2.1K]

Answer:

Far left one, A

Step-by-step explanation:

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3 years ago

Read 2 more answers

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nata0808 [166]

Answer:

Step-by-step explanation:

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3 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago