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3 years ago

8

Lin and Andre start walking toward each other at the same time from opposite ends of a 22 mile walking trail. I am to lazy to ty

pe the rest so read the image and fill out the chart please! Thank you!

2 answers:

notka56 [123]3 years ago

4 0

Answer:

2 hours- 5 miles, 6 miles, 13 miles

3 Hours- 7.5, 9, 9.5

4 hours- 10, 12, 6

Step-by-step explanation:

I think it might be wrong but I did my best sorry!

algol [13]3 years ago

3 0

Answer:

I think It is 4 hours

Step-by-step explanation:

because 22 - 16.5 =5.5 in one hour they can 5.5 miles. so 5.5 x 4 = 22

so 4 hours should be the answer.

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Liz flips a coin 70 times. The coin lands heads up 28 times and tails up 42 times. Complete each statement

sveta [45]

Answer:

39%

Step-by-step explanation:

you would divide the number of times the coin landed on heads by the total times thrown which give s you .3857 you move the decimal over twice to give you 38.5 and round up to 39%

3 0

3 years ago

Read 2 more answers

Pls help I really need to pass this if like for 4th quarter where I m pls help me

Tom [10]

Answer:

✨Both will be the same✨

✨Step-by-step explanation:✨

there is 1 in each section 61 and 38

HOPE THAT HELPS

3 0

3 years ago

Express these numbers in standard form (a)5730(b)0.23(c)0.000231(d)7150000(e)500.8

alisha [4.7K]

Answer:

5730 = 5.73×10³

0.23 = 2.3×10‐¹

0.000231 = 2.31×10-⁴

7150000 = 7.15×10⁶

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8 0

4 years ago

The difference between eight and twice a number, x, is four.

Alisiya [41]

Answer:

0

Step-by-step explanation:

8 - 2x

8 -2(4)

8 - 8=0

3 0

3 years ago

In Major League Baseball, the American League (AL) allows a designated hitter (DH) to bat in place of the pitcher, but in the Na

hammer [34]

Answer:

Step-by-step explanation:

From the given information:

Let:

$\mu_1$ represent the population mean no. for DH group.

$\mu_2$ represent the population mean no. for no DH group

$n_1$ represent the sample sizes of DH

$n_2$ represent the sample sizes of no DH

Sample size: $n_1 = n_2 = 20$

For both groups, the population standard deviation of runs scored = 2.54

i.e

$\sigma^2_1 = \sigma_2^2 = 2.54$

The null & alternative hypothesis;

$H_o : \mu_1 \le \mu_2 \\ \\ H_a: \mu_1 > \mu_2$

Level of significance:

The test statistics is:

$z = \dfrac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma_2^2}{n_2} }}$

where;

$\bar x_1$ = sample mean for DH group

$\implies \dfrac{1}{n_1} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(0+6+8+2+2+4+7+7+6+5+1+1+5+4+4+5+7+11+10+0) \\ \\ = \dfrac{92}{20}$

= 4.6

$\bar x_2$ = sample mean for no DH group

$\implies \dfrac{1}{n_2} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(3+6+2+4+0+5+7+6+1+8+12+4+6+3+4+0+5+2+1+4) \\ \\ = \dfrac{83}{20}$

= 4.15

Now:

$z = \dfrac{4.60- 4.15}{\sqrt{\dfrac{2.54^2}{20} + \dfrac{2.54^2}{20} }} \\ \\ = \dfrac{0.45}{\sqrt{0.32258 +0.32258 }} \\ \\ = \dfrac{0.45}{0.803219} \\ \\$

= 0.5602

Since the test is one-tailed by looking at that $H_a:$

The P-value = $P(Z > z)$

$\implies 1 - P(Z \le z) \\ \\ = 1 - P(Z \le 0.5602)$

Using Excel Function " =normdist(z)"; we have":

$= 1- 0.7123$

P-value = 0.2877

Decision rule: To reject $H_o$ , if $p-value < \alpha \ at \ 0.10$

Conclusion: SInce P = 0.2877 which is $> \ \alpha \ at \ 0.10.$

We fail to reject the $H_o$ and conclude that there is insufficient evidence to support the given claim that: $\text{more runs are scored in games for which DH is used.}$

8 0

3 years ago