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elena-s [515]
3 years ago
14

Find from first principle,the differential coefficient of;sin(ax+b)

Mathematics
1 answer:
astra-53 [7]3 years ago
3 0

Answer:

Step-by-step explanation:

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6. Richy has a bag of marbles. He gave 5/12 of
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Answer:

7/12

Step-by-step explanation:

12/12-5/12=7/12

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3 years ago
During a sale at a bookstore, Joseph buys a book at full price. He is given a 50 percent discount on a second book of equal or l
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Total cost of books was 15 + 10 or 25.  With the 50 % discount on the $10 book, the total is now $20. The total cost was reduced by (B) 20%.
 $5 off the $25 price is a 20% discount.
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3 years ago
What is -5.8+(-2.5)
Scilla [17]

-5.8 + (-2.5) = -8.3

since both the number are negative you simply add the numbers and use the negative sign.

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3 years ago
Write an equation of the line passing through the point (-6,-7)with slope=4.
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5 0
3 years ago
Read 2 more answers
Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg
Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let <em>X</em> = number of college women on a diet.

The probability of a woman being on diet is, P (X) = <em>p</em> = 0.70.

The sample of women selected is, <em>n</em> = 267.

The random variable thus follows a Binomial distribution with parameters <em>n</em> = 267 and <em>p</em> = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions (\hat p) follows a Normal distribution.  

The mean of this distribution is:

\mu_{\hat p} = p = 0.70

The standard deviation of this distribution is: \sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z

**Use the <em>z</em>-table for the probability.

P(\hat p \geq 0.75)=1-P(Z

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0
3 years ago
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