What is the area of the polygon

Angelica’s bouquet of dozen roses contains 5 white roses. The rest of the roses. What fraction of the bouquet is pink? There are

Gennadij [26K]

Answer:

7/12

Step-by-step explanation:

total: 12 roses

white roses: 5

pink roses: 7

fraction of pink roses = 7/12

4 0

3 years ago

1+000 +0000 - 1 what is the best answer

creativ13 [48]

Answer:

It's 0 I think I'm not sure (I'm sure it's 0 but That's what I always say to humble myself so)

Step-by-step explanation:

5 0

2 years ago

Answer this please.

Tems11 [23]

Answer:

2. absolute value of p - q

Explanation:

Because when you calculate distance, you always use absolute value.

Can I get a thx and brainliest?

7 0

3 years ago

Read 2 more answers

Given h(x) = 5x + 3. what is the value of h(-11)

alexira [117]

Answer:

h(-11)=-52

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago