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fenix001 [56]
2 years ago
10

Can someone help me please with these questions

Mathematics
2 answers:
Ad libitum [116K]2 years ago
7 0
What is the question? You only gave shapes with dimensions.
Do you want us to find volume? Or surface area??
Come on please, dont leave us hangin.

Com.ment me what your question is

Klio2033 [76]2 years ago
5 0
Plus plus lang po yan
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Find the value of x , if log₂ [log₅ (log₃x)] = 0
Lina20 [59]
\log_2 (\log_5( \log_3x)) = 0\\
D:x>0 \wedge \log_3x>0 \wedge \log_5(\log_3x)>0\\
D:x>0\wedge x>1 \wedge\log_3x>1\\
D:x>1 \wedge x>3\\
D:x>3\\
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3 0
3 years ago
Help due now pleaseee
DENIUS [597]

Answer:

d

Step-by-step explanation:

base² + Altitude² = Hypotenuse²

 x² + 9² = 13²

x² + 81 = 169

       x² = 169 - 81

      x² = 88

      x = \sqrt{88}

     x = \sqrt{2*2*2*11}=2\sqrt{2*11}\\\\x=2\sqrt{22}

6 0
3 years ago
Read 2 more answers
6^2 + x - x^2 for x=3 <br> help me pls
Stels [109]

Answer:

30

Step-by-step explanation:

Just insert the value of x into the equation.

x = 3 right?

6^2 + 3 - 3^2 because x = 3

36 + 3 - 9 = 39 -9 = 30

:D

4 0
3 years ago
7th grade | discover the pattern and continue
german

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Step-by-step explanation:

5 0
2 years ago
Assume that the population of human body temperatures has a mean of 98.6 degrees F and a standard deviation of 0.62 degrees F. I
dimulka [17.4K]

Answer:

0% probability of getting a mean temperature of 98.2 degrees F or lower.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

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.

In this problem, we have that:

\mu = 98.6, \sigma = 0.62, n = 106, s = \frac{0.62}{\sqrt{106}} = 0.06

Find the probability of getting a mean temperature of 98.2 degrees F or lower.

This is the pvalue of Z when X = 98.2. So

Z = \frac{X - \mu}{s}

Z = \frac{98.2 - 98.6}{0.06}

Z = -6.67

Z = -6.67 has a pvalue of 0.

So there is a 0% probability of getting a mean temperature of 98.2 degrees F or lower.

8 0
2 years ago
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