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3 years ago

Help me please i wish i can get somehelp

Mathematics

1 answer:

USPshnik [31]3 years ago

4 0

Answer:

29. $\frac{1}{3}$

30. a) 32

b) Yes, the ratio will change 1 : 2.

Step-by-step explanation:

Question 29.

Given that:

Number of gold medals won by France = 11

Number of silver medals won by France = 9

Number of bronze medals won by France = 13

To find:

The ratio of gold medals won to the total number of medals won by France.

Solution:

First of all, we need to find the total number of medals won by France.

Total number of medals won = 11 + 9 + 13 = 33

The required ratio can be found as:

$\dfrac{\text{Number of gold medals won}}{\text{Total number of medals won}}\\\Rightarrow \dfrac{11}{33}\\\Rightarrow \bold{\dfrac{1}{3}}$

Question 30:

Number of girls = 20

Ratio of boys to Ratio of girls = 3 : 5

a. Total number of students in the class?

First of all, we need to find the number of boys in the class then we will add the value to the number of girls to get the answer.

Let the number of boys = $x$

As per the ratio, we can write:

$\dfrac{3}{5 } = \dfrac{x}{20}\\\Rightarrow x = 12$

Total number of students = 20 + 12 = 32

b. Four number of boys and four girls are moved to other class.

Number of boys left = 12 - 4 = 8

Number of girls left = 20 - 4 = 16

New ratio = 8 : 16 = 1 : 2

Therefore, the ratio got changed to 1 : 2.

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Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of bra

Stella [2.4K]

Answer:

The 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =114.1$ represent the sample mean 1

$\bar X_2 =129.9$ represent the sample mean 2

n1=5 represent the sample 1 size

n2=2 represent the sample 2 size

$s_1 =5.08$ sample standard deviation for sample 1

$s_2 =5.37$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =114.1-129.9=-15.8$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_1 +n_2 -1=5+5-2=8$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=\pm 2.31$