HELPPPPPPPPPPPPPPPPP WITH THIS MATHHH.

Triangle PQR is transformed to triangle P′Q′R′. Triangle PQR has vertices P(8, 0), Q(6, 2), and R(−2, −4). Triangle P′Q′R′ has v

pav-90 [236]

lo siento de antemano! Necesito los puntos para hacer una publicación sobre mi amigo desaparecido. Ella desapareció y no puedo encontrarla. Ella se ha ido desde Halloween gracias por su comprensión. 8 9 028 los números son para covienes que estoy cosiendo el quistoion. Lo siento, nuevamente.

3 0

2 years ago

Answer the question in the picture

Schach [20]

The area of the semicircle is A = 1187.9 cm².

<h3>What is the area of a circle?</h3>

The area of a circle with a radius of r is A = πr².

Given that, the diameter of the semicircle is 55 cm.

The radius of the semicircle is,

r = 55/2

The area of a semicircle is given by,

A = (1/2)πr²

Substitute the values,

A = (1/2)π(55/2)²

A = 1187.9

Hence, the area of the semicircle is A = 1187.9 cm².

Learn more about the area of a circle:

brainly.com/question/22964077

#SPJ1

6 0

1 year ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ 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 conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ 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 less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't know the population deviation so 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".

Calculate the statistic

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

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ 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 conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

2 years ago

F(x)=5500(1.65) at what percent are the bacteria growing each hour

kodGreya [7K]

So exponenetial growth or decay

it goes like this
f(t)=P(1+r)^t
r=rate of decay or grouth
if r>0 then it is grouth by r
if r<0 then it is decay
i
P=initial amount
so
given
f(x)=5500(1.65)^t
f(x)=5500(1+0.65)^t
0.65>0 so it is growing by 65%

3 0

2 years ago

What is the answer to 701km-523km445m

mixas84 [53]

You convert it to meters first. So you get 701000m and 523445mYou subtract the values and get this: 177555

So your answer is 177555

Hope this helps :)

8 0

3 years ago