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3 years ago

()+()+()+()+()=30

1 answer:

3 0

<span>There are 5 boxes, where we have to put given numbers. Addition of 5 odd numbers always gives an odd number result. 30 is an even number. So there is no combination of given numbers which can give you 30 as a result.
</span>

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The exchange rate at a post office is £1 = 1.17 euros <br><br> how many euros will u get for £280

Sati [7]

Answer:

327.6 €

Step-by-step explanation:

you can solve with an equation

1£ : 1.17€ = 280£ : x

1 : 1.17 = 280 : x

x = 1.17 * 280 : 1

x = 327.6

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check

1 : 1.17 = 280 : 327.6

0.85 = 0.85

the answer is good

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or you can simpy solve with a multiplication

280 * 1.17 = 327.6

8 0

3 years ago

Stuck on this question. May I please have some help?

Rina8888 [55]

Answer:

So a point is (-3,-5) and the vertex is (-4,-3)

Step-by-step explanation:

This is in vertex form. Vertex form is y=a(x-h)^2+k where (h,k) is the vertex.

The vertex here is (-4,-3)... now just use a value of x to plug in (any value besides -4)

I will choice -3. This gives -2(-3+4)^2-3

f(-3)=-2(1)^2-3

f(-3)=-2-3

f(-3)=-5

So a point is (-3,-5) and the vertex is (-4,-3)

4 0

3 years ago

Read 2 more answers

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who en

AVprozaik [17]

Answer:

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

$z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057$

$p_v =P(Z>0.0057)=0.4977$

The p value is a very high value and using the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.

Step-by-step explanation:

1) Data given and notation

$X_{M}=23$ represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

$X_{W}=8$ represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

$n_{M}=66$ sample of male selected

$n_{W}=23$ sample of demale selected

$p_{M}=\frac{23}{66}=0.34848$ represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

$p_{W}=\frac{8}{23}=0.34782$ represent the proportion of women with red/green color blindness

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:

Null hypothesis: $p_{M} \leq p_{W}$

Alternative hypothesis: $p_{M} > p_{W}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

Where $\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{23+8}{66+23}=0.34831$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057$

4) Statistical decision

Using the significance level provided $\alpha=0.05$ , the next step would be calculate the p value for this test.

Since is a one side right tail test the p value would be:

$p_v =P(Z>0.0057)=0.4977$

So the p value is a very high value and using the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.

4 0

3 years ago

Santo made 4 less than twice as many cupcakes as Lynn. Write an equation to represent the number of cupcakes that Santo made.

pickupchik [31]

Santo made 4 less than twice as many cupcakes as Lynn.

x * 2 - 4

5 0

3 years ago

Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o

Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is $\hat p_1$ = 1020/1272 = 85/106 = $0.8\overline {0188679245283}$ ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is $\hat p_2$ = 5343/6238 ≈ 0.8565

c) We have the following two way table;

$\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}$

$The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780$

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

$95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}$

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

$95 \% \ CI \approx 0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155$

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0

3 years ago