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Marrrta [24]
3 years ago
14

Lexi Susie and Ryan are playing on online word game Royals score 100034 points Lexi scores 9348 fewer points t

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0
100,435 hope this helps you
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A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Minchanka [31]
<h2><u>Answer with explanation</u>:</h2>

Let \mu be the distance traveled by deluxe tire .

As per given , we have

Null hypothesis : H_0 : \mu \geq50000

Alternative hypothesis : H_a : \mu

Since H_a is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.

Test statistic : z=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

where, n= sample size

\overline{x}= sample mean

\mu= Population mean

s=sample standard deviation

For n= 31,\ \sigma=8000,\ \overline{x}=46,800\ \&\ s=46,800, we have

z=\dfrac{46800-50000}{\dfrac{8000}{\sqrt{31}}}=-2.23

By using z-value table,

P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23)   [∵P(Z≤-z)=1-P(Z≤z)]

=1-0.9871=0.0129

 Decision : Since p value (0.0129) < significance level  (0.05), so we reject the null hypothesis .

[We reject the null hypothesis when p-value is less than the significance level .]

Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.

4 0
3 years ago
I need help on this question
egoroff_w [7]
25 is the answer hope it helped
8 0
3 years ago
If y'all can answer any of these I'll be thankful :)
AURORKA [14]
Sorry we can’t help with this
4 0
3 years ago
Apply Newton’s method to estimate the solution of x3 − x − 1 = 0 by taking x1 = 1 and finding the least n such that xn and xn +
ss7ja [257]

Solution :

Given

$f(x)=x^3-x-1, x_1=1$

$f'(x)=3x^2-1$

Let the initial approximation is $x_1 =1$

So by Newton's method, we get

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$

$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$

$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$

$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$

$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$

$x_5 \approx x_6$ are identical up to eight decimal places.

The approximate real root is x ≈ 1.32471795

∴ x = 1.32471795

4 0
3 years ago
There are 2 squares and 8 circles. What is the simplest ratio of squares to total shapes?​
Veronika [31]

Answer:

1/4

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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