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vagabundo [1.1K]

3 years ago

7

A Treasure hunt competition has eliminations each round. The table below shows the number of players in each of the first 5 roun

ds of the tournament.

1 answer:

vovikov84 [41]3 years ago

5 0

I think its c if i did it right

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What is the greatest whole number that rounds to 54’300

ipn [44]

If that is 54.300 the greatest whole number would be 54 if it is supposed to be 54,300 then it would stay the same and be 54,300

5 0

4 years ago

Read 2 more answers

How do you do<br>8(4x+7)=56

lilavasa [31]

Multiply 8 to 4x and 7 so it looks like 32x +56=56 then you subtract 56 from to 56 which equals zero. Your equation should look like 32x=0. Divide 32 by 0 which equals 0. X=0

4 0

4 years ago

Given that sinθ=8/17 , what is cosθ ? Enter your answer, as a simplified fraction, in the boxes.

leonid [27]

Cosθ = 15/17

Step-by-step explanation:

Sinθ = 8/17

We know that,

Sin^2 θ + cos^2 θ =1

64/289 + cos^2 θ =1

1 - 64/289 = cos^2 θ

225/289 = cos^2 θ

Cos θ = √225/289

Cos θ = 15/17

4 0

3 years ago

From 309 grams to 299 grams.

Tom [10]

Answer: 3 percent decrease. = 0% Increase or a 5 % decrease.

Step-by-step explanation: Sorry, I don't have an explanation!

3 0

4 years ago

Based on an indication that mean daily car rental rates may be higher for Boston than for Dallas, a survey of eight car rental c

Elina [12.6K]

Answer:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

Step-by-step explanation:

Data given

$n_1 =8$ represent the sample size for group Boston

$n_2 =9$ represent the sample size for group Dallas

$\bar X_1 =47$ represent the sample mean for the group Boston

$\bar X_2 =44$ represent the sample mean for the group Dallas

$s_1=3$ represent the sample standard deviation for group Boston

$s_2=3$ represent the sample standard deviation for group Dallas

We can assume that we have independent samples from two normal distributions with equal variances and that is:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

Let the subindex 1 for Boston and 2 for Dallas we want to check the following hypothesis:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t student distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Replacing we got:

$\S^2_p =\frac{(8-1)(3)^2 +(9 -1)(3)^2}{8 +9 -2}=9$

And the deviation would be just the square root of the variance:

$S_p=3$

The statitsic would be:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

5 0

4 years ago