Sam has eight shells and thandeka has 12 shells .what is the ratio ,in simplest form ,of Sam's shells to thandeka's shells?

SHOW YOU WORK AND EXPLAIN PLS I WILL MARK YOU BRAINLIEST PLEASE.

KatRina [158]

Answer:

a) 1.8 $cm^2$

b) 3.6 $feet^2$

Step-by-step explanation:

Conversion and calculation of areas

The area of a rectangle is A=wh, where w is the width and h is the height. There are 3 feet in one yard

a)

We are told Kalie put 6 stickers, each one of 1/2 centimeters (0.5 cm) wide by 3/5 (0.6 cm) centimeter long. The area of one sticker is

A=(0.5)(0.6)=0.3 $cm^2$

Assuming there is no overlapping, the 6 stickers have a total area

6*0.3 $cm^2$ =1.8 $cm^2$

b)

Each of Elana's wrapping papers measures 2/5 yards long and 1/4 yard wide. Converting them to feet we have

long=2/5*3=1.2 feet

wide=1/4*3=0.75 feet

Area of each paper=1.2 feet*0.75 feet=0.9 $feet^2$

Area of the entire board, assuming no overlapping and no space left uncovered=4*0.9 $feet^2$

Area of board=3.6 $feet^2$

7 0

3 years ago

QUESTIONNNNNNNNNN!!!!

Lemur [1.5K]

Answer:

4 i think

Step-by-step explanation:

3 0

2 years ago

If an item is marked up 15% and the new cost is 155.25 what was the original price

Dmitry [639]

(1-0.15)=0.85
0.85 (155.25)=131.962
estimated answer
131.962 is about $132.00

4 0

3 years ago

Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of i

kolezko [41]

Answer:

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

$p_v =2*P(t_{4}>6.943)=0.00226$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.

Step-by-step explanation:

Data given and notation

Data: 801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=811$ represent the sample mean

$s=19.647$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =750$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

What do you conclude?

Compute the p-value

Since is a two tailed test the p value would be:

$p_v =2*P(t_{4}>6.943)=0.00226$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.

4 0

3 years ago

At a large company banquet for several thousand employees and their families, many of the attendees became ill the next day. The

yan [13]

Answer:

a. Yes. This provides convincing evidence that the true proportion of all attendees who ate the fish that got sick (80%) is more than the true proportion of all attendees who did not eat the fish that got sick.

b. The mistake here would have been the rejection of the Doctor's theory or hypothesis to the effect that more attendees who ate the fish got sick than those who did not eat the fish. This is a Type 1 error. A Type 1 error occurs when a null hypothesis is rejected when it is true. On the other hand, a Type II error occurs when the null hypothesis is accepted when it should be rejected. While a Type I error is equivalent to a false positive, a Type II error is equivalent to a false negative.

Step-by-step explanation:

Total number of attendees who ordered fish = 1,000

Sample size of the attendees who ate fish = 80

Number of attendees who ate the fish and got sick = 64 (80% or 64/80)

Sample size of attendees who did not eat fish = 60

Number of attendees who did not eat fish and got sick = 39 (65% or 39/60)

4 0

3 years ago