Witch fraction is equivalent to 33/200

I NEED HELP ASAP PLEASEEEEEEEE!!!!!!!!!!

igomit [66]

C

Step-by-step explanation:

In my opinion by looking at the photograph I think its 6 in my opinion sorry if I'm wrong

7 0

3 years ago

The students at Norton School were asked to name their favorite type of pet. Of the 430ilar results, about students surveyed, 25

elena-s [515]

Answer: 60

Step-by-step explanation:

From the question, we are informed that 430 students were surveyed and 258 said that their favorite type of pet was a dog. The fraction/decimal of those that chose dog as their favorite pet will be:

= 258/430

= 3/5

= 0.6

Suppose that only 100 students were surveyed, the number of students that would say that a dog is their favorite types of pet will be:

= 0.6 × 100

= 60

5 0

3 years ago

Polygon C and polygon D are similar. The perimeter of polygon C is 192 inches, and the perimeter of polygon D is 48 inches. If o

Marianna [84]

Answer:

B. 2 in.

Step-by-step explanation:

First, put the two perimeters together to form a fraction.

192/48, which can simplify to 4/1

Now, following the order in which you put the perimeters (I used C/D,) set it equal to the sides.

$\frac{4}{1} = \frac{8}{x}$

Now, cross multiply. (multiply 1 and 8, 4 and x.)

$4x = 8$

Now, solve.

x=2 inches.

3 0

3 years ago

Can somebody help me, question down below <3

taurus [48]

It’s C! they distributed 12 to x and 4 to get 12x+48

5 0

3 years ago

Read 2 more answers

When people make estimates, they are influenced by anchors to their estimates. A study was conducted in which students were aske

12345 [234]

Answer:

The null and alternative hypothesis are:

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

where μ1: mean calorie estimation for the cheesecake group and μ2: mean calorie estimation for the organic salad group.

There is enough evidence to support the claim that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first (P-value=0.0000002).

Step-by-step explanation:

The question is incomplete:

"Suppose that the study was based on a sample of 20 people who thought about the cheesecake first and 20 people who thought about the organic fruit salad first, and the standard deviation of the number of calories in the cheeseburger was 128 for the people who thought about the cheesecake first and 140 for the people who thought about the organic fruit salad first.

At the 0.01 level of significance, is there evidence that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first?"

This is a hypothesis test for the difference between populations means.

The claim is that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first.

Then, the null and alternative hypothesis are:

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

The significance level is 0.01.

The sample 1 (cheese cake), of size n1=20 has a mean of 780 and a standard deviation of 128.

The sample 2 (organic salad), of size n2=20 has a mean of 1041 and a standard deviation of 140.

The difference between sample means is Md=-261.

$M_d=M_1-M_2=780-1041=-261$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{128^2}{20}+\dfrac{140^2}{20}}\\\\\\s_{M_d}=\sqrt{819.2+980}=\sqrt{1799.2}=42.417$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-261-0}{42.417}=\dfrac{-261}{42.417}=-6.153$

The degrees of freedom for this test are:

$df=n_1+n_2-1=20+20-2=38$

This test is a left-tailed test, with 38 degrees of freedom and t=-6.153, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.0000002) is smaller than the significance level (0.01), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first.

3 0

3 years ago