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Nadusha1986 [10]

3 years ago

6

Apersoncanpay $ 6 foramembershiptotheartmuseumandthengotothemuseumforjust $ 1 pervisit. Whatisthemaximumnumberofvisitsamemberoft

heartmuseumcanmakeforatotalcostof $ 42 ?

1 answer:

labwork [276]3 years ago

6 0

Now I'm not completely sure of this answer but by my calculations the member can visit at least 36 times with $42.

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If (6^0)^x= 1 what are the possible outcomes of x? Explain your answer.

swat32

X can be any real number

As long as it is to the power of 0, it will equal one. And because 6^0 = 1, 1^x can be any positive number, and it will still give you one.

As stated by WojtekR, even negative numbers can get 1

x = - 5

(6^0)^-5 = 1^-5 = 1/1 = 1

~ Thanks @WojtekR

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3 years ago

What is the standard form of this circle m(2,4) n(9,4)

Helga [31]

You'd have to find slope so you would use y1-y2 over x2-x1 so itd be 4-4 over 9-2 thatd equal 0/7 thats ur slope. using slope intercept form y=mx+b plug in ur slope y=0/7 x + 4 (y- intecept) then youd subtract the 0/7 x from both so youd have -0/7x+y=4 and that is in standard form.

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3 years ago

What is the range of the function on the graph

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3 years ago

Radioactive fallout from testing atomic bombs drifted across a region. There were 220 people in the region at the time and 36 of

vazorg [7]

Answer:

(a) Yes, the death rate observed in the group unusually high.

(b) Yes, this prove that exposure to radiation increases the risk of cancer.

Step-by-step explanation:

In a region where radioactive fallout from testing atomic bombs drifted across, 36 of the 220 people died of cancer.

The sample proportion of the number of people dying of cancer in this region is:

$\hat p =\frac{36}{220} =0.164$

It is estimated by the cancer expert that there will be 33 cancer deaths in a group of this size.

The population proportion is: $p=\frac{33}{220} =0.15$

A hypothesis test for single proportion can be performed to test if the proportion of cancer deaths was high or not and whether it was due to the increase in radiation.

The hypothesis can be defined as:

<em>H</em>₀: The proportion of death due to cancer in this region is 0.15, i.e. <em>p</em> = 0.15.

<em>H</em>ₐ: The proportion of death due to cancer in this region is more than 0.15, i.e. <em>p</em> > 0.15.

Assuming the population is Normally distributed since the sample size is large.

Assume the level of significance is <em>α</em> = 0.05.

The test statistic is:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}\\= \frac{0.164-0.15}{\sqrt{\frac{0.15(1-0.15)}{220}}} \\=0.581$

The test statistic value is 0.581.

<u>Decision rule:</u>

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis is rejected and vice versa.

The <em>p</em>-value of the test is:

$P(Z>0.581)=1-P(Z$

**Use a <em>z</em>-table for the <em>p</em>-value.

The <em>p-</em>value = 0.281 > <em>α</em> = 0.05.

The null hypothesis was failed to be rejected.

(a)

As the null hypothesis was not rejected at 5% level of significance it can be concluded that the death rate due to cancer observed in this group of people is unusually high.

Yes, the death rate observed in the group unusually high.

(b)

It was previously mentioned that a radioactive fallout from testing atomic bombs drifted across this region. Due to this 36 people died of cancer in this region.

As the death rate observed in the group unusually high, it can be said that the increase in the death rate due to cancer was due to the excessive exposure to radiation.

Yes, this prove that exposure to radiation increases the risk of cancer.

5 0

3 years ago

The line segment joining the points a (3,2) and b (5,1) is divided at the point p in the ratio 1:2 and p lies on the line 2x-18y

kumpel [21]

your answer would be 2x-18y+k=0

8 0

3 years ago