This was number 19 and 20 sry guuyyyyysssss

Are the two triangles similar? If so, state the reason and the similarity statement.

joja [24]

$\displaystyle\\ \frac{AY}{ED}=\frac{8}{4}=2\\\\\frac{AM}{EC}=\frac{6}{3}=2\\\\ \frac{YM}{DC}=\frac{10}{5}=2\\\\\implies k = 2\\\\\text{Correct answer: } \boxed{\bf Yes; SSS; \Delta AYM \sim \Delta EDC}$

7 0

3 years ago

Read 2 more answers

A coin is tossed 50 times, and the number of times heads comes up is counted. Which of the following statements about the distri

TEA [102]

Answer:

based on the fact there are no listed choices, a true statement would be about how the number of times heads came up would be about 50% or 25 times

Step-by-step explanation:

5 0

2 years ago

Luis wants to buy a skateboard that usually sells for $79.97. All merchandise is discounted by 12%. What is the total cost of th

Tju [1.3M]

Answer:

$74.8

Step-by-step explanation:

step 1: apply the discount to the price

12% = .12

$79.97 X .12 = $9.5964

$79.97 - $9.5964 = $70.3736

step 2: apply tax

$70.3736 X .0625 = $4.39835

$70.3736 + $4.39835 = $74.77195

step3: round

$74.8

5 0

3 years ago

A soft-drink machine at a steakhouse is regulated so that the amount of drink dispensed is normally distributed with a mean of 2

AnnZ [28]

Answer:

$z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6$

$p_v =P(z>0.6)=0.274$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=203$ represent the mean height for the sample

$\sigma=15$ represent the population standard deviation

$n=9$ sample size

$\mu_o =200$ represent the value that we want to test

$\alpha=0.05$ 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.

We need to conduct a hypothesis in order to check if the mean is higher than 200, the system of hypothesis would be:

Null hypothesis: $\mu \leq 200$

Alternative hypothesis: $\mu > 20$

If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

Calculate the statistic

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

$z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6$

P-value

Since is a one sided test the p value would be:

$p_v =P(z>0.6)=0.274$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.

8 0

3 years ago

Danielle wants to know if a new dresser will fit next to her bed. The dresser she would like to buy is 73 centimeters wide. She

zhannawk [14.2K]

Answer:

No.

The dresser's width(28.74in) is greater than the remaining free width(10in) in the room.

Step-by-step explanation:

First, convert the chest width into inches, $73cm=28.74 \ in$

To find the excess width after fitting the bed in her room, we subtract the bed width from the room's width:

$Width_R=86 \ in\\Width_B=76 \ in\\\\Width_{excess}=Width_R-Width_B\\\\=86\ in -76\ in\\\\=10\ in$

Since 10in is less than 28.74in, the dresser will not fit next to her bed.

5 0

3 years ago