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3 years ago

Please help me(ik it’s not C)

Mathematics

2 answers:

Irina-Kira [14]3 years ago

8 0

I think the correct answer is B :)

Gemiola [76]3 years ago

7 0

It is B :)) that’s the answer

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Helphelphelphelphelphelppppppp

n200080 [17]

Answer:

Step-by-step explaSimplify the equation by finding the square root of both sides. √x2=x √0=0. x=0. Check: 02=0.

5 0

3 years ago

A manufacturer has fixed costs of $16,000 per month and can produce w widgets at a cost of 0.1w^(2) + 20w. How many widgets shou

musickatia [10]

Consider c as the cost of the widget so that our given equation is
c = 0.1w^2 + 20w
Take the derivate of the equation.
d/dt (c = 0.1w^2 + 20w)
dc/dt = 0.2w + 20
Given dc/dt = $16000 per month, the number of widgets would contain:
16000 = 0.2w + 20
-0.2w = 20 - 16000
-0.2w = -15980
w = 79900 widgets

5 0

3 years ago

What’s the answer to this

Ipatiy [6.2K]

2.4/0.3=8
0.3x8=2.4
The scale factor is 8.

7 0

3 years ago

According to data from the state blood program, 40 percent of all individuals have group A blood. If six individuals give blood,

Serjik [45]

Answer: 0.27648

Step-by-step explanation:

Given : The proportion of all individuals have group A blood : p=0.040

Total individuals give blood : n= 6

Let X be the number of individuals have group A blood.

Since all individual are independent of each other.

$X\sim Bin(n=6, p=0.40)$

Formula : $P(X=x)=^nC_xp^x(1-p)^{n-x}$ , where n= sample size , p = probability of getting success in each trial.

The probability that exactly three of the individuals have group A blood. :

$P(x=3)= ^6C_3(0.40)^3(1-0.4)^3\\\\= \dfrac{6!}{3!3!}\times(0.40)^3(0.60)^3\\\\=0.27648$

The probability that exactly three of the individuals have group A blood. is 0.27648

8 0

4 years ago

If the distribution is really (5.43,0.54)

defon

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be 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.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

$\mu = 5.43, \sigma = 0.54$

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

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

$Z = \frac{5.79 - 5.43}{0.54}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

7 0

4 years ago