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Natalija [7]
3 years ago
6

15 points please explain thanks

Mathematics
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

y=29

Step-by-step explanation:

The angles are vertical angles so they are equal.

4y-8 = 79+y

Subtract y from each side

4y-y -8 = 79+y-y

3y -8 = 79

Add 8 to each side

3y-8+8 = 79+8

3y = 87

Divide each side by 3

3y/3 = 87/3

y=29

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Suppose that you have an enormous grapefruit that is 92% water (by weight). The grapefruit weights 100 pounds. If the water cont
yuradex [85]

Answer:

The weight of grapefruit is now 80 pound.

Step-by-step explanation:

Consider the provided information.

Let the x is the weight loss. The weight of grapefruit is 100 pounds and water is 92%. After evaporation water is 90%.

Thus the weight loss is:

0.92\times100-0.90(100 - x) = x

92-90+0.90x=x

2=x-0.90x

2=0.1x

x=20

Hence, the weight loss is 80 pounds.

Therefore,  New weight is 100 - 20 = 80 pounds

The weight of grapefruit is now 80 pound.

3 0
3 years ago
What is the 15th term of the sequence -4, -2, 0, 2, …?What is the 15th term of the sequence -4, -2, 0, 2, …?
Dmitriy789 [7]

Answer:

24

Step-by-step explanation:

6 0
3 years ago
Try some calculations using the keypad or your keyboard
morpeh [17]

Answer:

the -3.2+8.1 is 4.9 I can't see the rest

6 0
3 years ago
Each item produced by a certain manufacturer is independently of acceptable quality with probability 0.95. Approximate the proba
Diano4ka-milaya [45]

Answer:

The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

Step-by-step explanation:

Let <em>X</em> = number of items with unacceptable quality.

The probability of an item being unacceptable is, P (X) = <em>p</em> = 0.05.

The sample of items selected is of size, <em>n</em> = 150.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and <em>p</em> = 0.05.

According to the Central limit theorem, if a sample of large size (<em>n</em> > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.

The mean of this sampling distribution is: \mu_{\hat p}= p=0.05

The standard deviation of this sampling distribution is: \sigma_{\hat p}=\sqrt{\frac{ p(1-p)}{n}}=\sqrt{\frac{0.05(1-.0.05)}{150} }=0.0178

If 10 of the 150 items produced are unacceptable then the probability of this event is:

\hat p=\frac{10}{150}=0.067

Compute the value of P(\hat p\leq 0.067) as follows:

P(\hat p\leq 0.067)=P(\frac{\hat p-\mu_{p}}{\sigma_{p}} \leq\frac{0.067-0.05}{0.0178})=P(Z\leq 0.96)=0.8315

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

Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

5 0
3 years ago
C²_4<br> Which expression is equivalent to C+3<br> C+2<br> 3(2-0)
Arturiano [62]

Answer:

<h2>its equivalent to , x (12 y+4)</h2><h2 />

6 0
3 years ago
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