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Angelina_Jolie [31]
3 years ago
10

If middle school and senior school are winning football games in the ratio 3:9 . If the Middle school won 17 games, calculate ho

w many games did the senior school win?
Mathematics
1 answer:
Helen [10]3 years ago
4 0
The Middle school won 17 games so the senior school won 51 games

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Triangle A'B'C' is formed using the translation (x + 1. y + 1) and the dilation by a scale factor of 3 from the origin. Which eq
ArbitrLikvidat [17]

Answer:

d

Step-by-step explanation:

(x, y) ⇒ (x + 1, y +1) ⇒ (3x, 3y)

A (-3, 3) ⇒ A' (-2, 4) ⇒ A'' (-6, 12)

B (1, -3) ⇒ B' (2, -2) ⇒ B'' (6, -6)

C (-3, -3) ⇒ C' (-2, -2) ⇒ C'' (-6, -6)

8 0
3 years ago
Question 1 options:Residents in Portland, Oregon think that their city has more rainfall than Seattle, Washington. To test this
dimaraw [331]

Answer:

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0

P-value = 0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

Step-by-step explanation:

We have to test the hypothesis of the difference between means.

The claim is that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0

The significance level is 0.10.

The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.

The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.

The difference between means is:

M_d= M_1-M_2=37.50-37.07=0.43

The standard error for the difference between means is:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774

We can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393

The degrees of freedom are:

df=n1+n2-2=45+35-2=78

Then, the p-value for this one-tailed test with 78 degrees of freedom is:

P-value=P(t>1.1393)=0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

7 0
3 years ago
Elmer has a collection of 300 fossils. Of these, 21% are fossilized snail shells.
sergeinik [125]

Answer:

63

Step-by-step explanation:

You need to find 21% of 300.

To find a percent of a number, multiply the percent by the number. Change the percent to a decimal by moving the decimal point of the percent two places to the left.

21% of 300 =

= 21% * 300

= 0.21 * 300

= 63

6 0
4 years ago
How to calculate the percentage decrease in third year.i want extra detailed working
kolezko [41]
If you want detailed working then I am not the right person. c:
8 0
4 years ago
Please help and no guessing please I need this.
vodka [1.7K]

Answer:

1. n + 8 = 10

2. \frac{n}{12}= 10

3. 8n - 6 = 34

4. 3(n + 12) = 45

5. No

6. Yes

Step-by-step explanation:

(only for questions 5 and 6)

5. If you substitute in y = 1, you get 1 * 49 = 0 or 49 = 0 which is not true.

6. If you substitute in b = 8, you get:

5(8-3)= 25\\5(5) = 25\\25 = 25

this is true.

7 0
3 years ago
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