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3 years ago

Need help with question please

Mathematics

1 answer:

viktelen [127]3 years ago

6 0

Answer:

The answer is x = -7.7 ans x = -1.3

Step-by-step explanation:

Here is the work I did. I used the quadratic formula. Sorry about my bad handwriting.

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Solve each question using square root property (5x-1)^2=16

Goshia [24]

Answer:

(5x-1)^2=16

Step-by-step explanation:

7 0

3 years ago

3.2y-1.4y+y-0.6y=0.55

Kazeer [188]

Answer:

it is y=0.25

Step-by-step explanation:

because Step 1: Simplify both sides of the equation.

3.2y−1.4y+y−0.6y=0.55

3.2y+−1.4y+y+−0.6y=0.55

(3.2y+−1.4y+y+−0.6y)=0.55(Combine Like Terms)

2.2y=0.55

Step 2: Divide both sides by 2.2.

2.2y

2.2

0.55

2.2

8 0

3 years ago

Here is some information about a play starts at 7:30pm first act last 46 minutes interval lasts 20 minutes second act last 53 mi

Yuki888 [10]

Answer:

9:29 PM

Step-by-step explanation:

Let us add the number of minutes the first act, interval and second act lasted:

46 + 20 + 53 = 119 minutes

Converting this to hours and minutes:

60 minutes = 1 hour

119 minutes = 1 hour 59 minutes

Therefore, the time that the second act ends is:

H M

7 : 30 PM

+ 1 : 59

9 : 29

The second act will end by 9:29 PM

8 0

3 years ago

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago

Please answer! It is due in 20 min.

White raven [17]

Answer:

Step-by-step explanation:

90-57=?

57+?=90

57+33=90

4 0

2 years ago