Write the fraction or mixed number as a percent.<br> 1 21/50 =

Evaluate ( 4 upon 3 ) -2<br><br><br>please amswer fast

Taya2010 [7]

<h2><u>Solution</u></h2>

$\frac{4}{3} - 2 \\ \\ = \frac{4 - 6}{ 3} \\ \\ = \frac{ - 2}{3}$

<h3>Hope This Helps You ❤️</h3>

6 0

3 years ago

If you flip the five coins, and the probability of one coin being "tails" is 0.50, what is the probability of getting "tails" on

jenyasd209 [6]

Answer: 0.03125

Step-by-step explanation:

We know that the probability of getting a tail , we toss a fair coin = 0.5

Given : Total number of trials = 5

Using binomial probability formula :

$P(x)=^nC_xp^x(1-p)^{n-x}$ , where P(x) is the probability of getting success in x trails, n is total number of trials and p is the probability of getting success in each trial.

The probability of getting "tails" on all five coins :_

$P(x=5)=^5C_5(0.5)^5(1-0.5)^0=(1)(0.5)^5=0.03125$

Hence, the probability of getting "tails" on all five coins =0.03125

8 0

3 years ago

The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.

Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If $Z \leq -2$ or $Z \geq 2$ , the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1$

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{19 - 22}{1}$

$Z = -3$

$Z = -3 \leq -2$ , so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0

2 years ago

On a standard reticulocyte preparation with new methylene blue, there are 100 cells counted with blue-stained granulofilamentous

DIA [1.3K]

Answer:

0.322 × 10¹² /L

Step-by-step explanation:

Data provided in the question:

Number of cells counted with blue-stained granulofilamentous material

i.e number of RET = 100

The red blood cells count = 3.22 × 10¹² /L

Hematocrit = 30%

Now,

RET% = [ [ Number of RET ] ÷ 1000 RBCs ] × 100%

= [ 100 ÷ 1000 ] × 100%

= 0.1 × 100%

= 10%

also,

Absolute reticulocyte count = ( %RET × RBC count ) ÷ 100

= [ 10 × 3.22 × 10¹² /L ] ÷ 100

= 0.322 × 10¹² /L

8 0

3 years ago

A savings account balance is compounded continuously.If the interest rate is 3.1% per year and the current balance is 1077.00 in

Masteriza [31]

Answer:13.8 is the best answer i could get

Step-by-step explanation:1,077 multiply that × 13.8 =14,862

8 0

3 years ago

Write the fraction or mixed number as a percent. 1 21/50 =