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4 years ago

80 POINTS!

Mathematics

2 answers:

Aleonysh [2.5K]4 years ago

7 0

4y Squared. Turns to 2y * 2y. Another way is 2*2*y*y

Margaret [11]4 years ago

3 0

4y².

You can rewrite this as 2y * 2y or 4 * y * y

You might be interested in

which experimental probability from coach nelson's experiment equals the theorectical probability? what is this probability?

Lilit [14]

Answer:

QUESTION:

which experimental probability from coach nelson's experiment equals the theorectical probability? what is this probability?

ANSWER:

This will explain everything to you.

https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-probability-statistics/cc-7th-theoretical-and-experimental-probability/v/comparing-theoretical-to-experimental-probabilites

Step-by-step explanation:

Hope that this helps you out! :)

If you have any questions please put them in the comment section below this answer.

Have a great rest of your day/night!

Please thank me on my profile if this answer has helped you.

7 0

3 years ago

Label each tick mark with its location on the number line:

Aleksandr-060686 [28]

From my understanding the question is basically asking u to label each Dash/tick mark on the line. They have already given u 0 as the first one, 24 as the fifth one and 60 as the eleventh one. I’m assuming u can use any method.
Answer: use any method of division or skip counting then you will see that each line increases by 6

First line: 0
Second line: 6
Third line: 12
Fourth line: 18
Fifth line: 24
Sixth line: 30
Seventh line: 36
Eight line: 42
Ninth line: 48
Tenth line: 54
Eleventh line: 60

4 0

3 years ago

Read 2 more answers

Nt Addition and Cogs tru D 13 F.

Serga [27]

Answer:

666665555

Step-by-step explanation:

nasnfojn jkfnsdjjdsbfbdsf

7 0

3 years ago

Please explain how to do this and still get the answer 16. Thank you!

Kitty [74]

(j-3)8=104
8j-24=104
8j=128
j=16

5 0

3 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were born, a

Masja [62]

Answer:

(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Yes, the proportion of girls is significantly different from 0.50.

Step-by-step explanation:

We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 400 babies were born, and 340 of them were girls.

(a) Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

p = population percentage of girls born

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ , $0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ ]

= [0.804 , 0.896]

Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Let p = population proportion of girls born.

So, Null Hypothesis, $H_0$ : p = 0.50 {means that the proportion of girls is equal to 0.50}

Alternate Hypothesis, $H_A$ : p $\neq$ 0.50 {means that the proportion of girls is significantly different from 0.50}

The test statistics that will be used here is One-sample z proportion test statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

So, the test statistics = $\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }$

= 19.604

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of girls is significantly different from 0.50.

8 0

3 years ago