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uysha [10]
3 years ago
13

Pls help meeeeeeeee will give BRAINLIEST

Mathematics
2 answers:
inn [45]3 years ago
8 0

Answer:

It's d) 4r+13b+16

Step-by-step explanation:

Hope this helps!

Amiraneli [1.4K]3 years ago
6 0
D.) 4r+13b+16

hope this helps. :)
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Consider the function f(x)=x3+6x2−20x+450.
Damm [24]

Answer:

Therefore reminder = 2802

Step-by-step explanation:

f(x)=x³+6x²-20x+450

x-12)x³+6x²-20x+450(x²+18x+196

      x³-12x²

    ____________________

          +18x²-20x+450

           18x²-216x

          _______________

                  196x +450

                  196x-2352

                 _____________

                            2802

Therefore reminder = 2802

6 0
3 years ago
Which is greater 6.76 or6.759
vaieri [72.5K]

Answer:

6.76

Step-by-step explanation:


7 0
3 years ago
Read 2 more answers
8.5=6.5(2d-3)+b can you help me to awarie it
inessss [21]

Answer:

b=-13d+28

Step-by-step explanation:

that's the answer friend

3 0
3 years ago
How many data entry clerks would you hire for a day if you spend 3000?
egoroff_w [7]
You would be able to hire 40.  Hope this helps
4 0
4 years ago
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A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken
mihalych1998 [28]

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

X=1 number of citrus trees that are infected with the citrus red mite.

n=5 random sample taken

\hat p=\frac{1}{5}=0.2 estimated proportion of citrus trees that are infected with the citrus red mite.

p true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

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{\hat p(1-\hat p)}{n}})

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values would be given by:

t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.2(1-0.2)}{(\frac{0.01}{1.96})^2}=6146.56  

And rounded up we have that n=6147

5 0
4 years ago
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