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3 years ago

8

The school cafeteria serves pizza every sixth day and applesauce every eighth day. Suppose that pizza and applesauce are both on

today's menu. In how many days will they be together on the menu again?

1 answer:

beks73 [17]3 years ago

3 0

They will be served again on the same day in 48 days

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Benjamin bought 12 goldfish. Each goldfish cost $0.98. How much did Benjamin spend

jeka57 [31]

If each goldfish cost $0.98 simply multiply this by the amount of fish bought:

(0.98)(12)
= 11.76

Benjamin spent $11.76 on his goldfish

Hope this helped!

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3 years ago

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Can yall help me out ?

larisa86 [58]

I thinks it’s line 5

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3 years ago

I don't this question so I need help with this problem

sveta [45]

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3 years ago

Can someone help me on this question?

notsponge [240]

Answer:

x = 19.6

Step-by-step explanation:

By applying cosine rule in ΔBCD,

cos(30)° = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

= $\frac{BC}{BD}$

$\frac{\sqrt{3}}{2}=\frac{BC}{16}$

BC = $8\sqrt{3}$

Now, by applying sine rule in ΔBAC,

sin(45)° = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\frac{1}{\sqrt{2}}=\frac{AB}{BC}$

$\frac{1}{\sqrt{2}}=\frac{x}{8\sqrt{3}}$

x = $8\sqrt{6}$

x = 19.59

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3 0

3 years ago

In 1943, an organization surveyed 1100 adults and asked, "Are you a total abstainer from, or do you on occasion consume, alc

Fynjy0 [20]

Answer:

We conclude that the proportion of adults who totally abstain from alcohol has changed.

Step-by-step explanation:

We are given that in 1943, an organization surveyed 1100 adults and asked, "Are you a total abstainer from, or do you on occasion consume, alcoholic beverages?"

Of the 1100 adults surveyed, 429 indicated that they were total abstainers. In a recent survey, the same question was asked of 1100 adults and 352 in

<u><em>Let p = proportion of adults who totally abstain from alcohol.</em></u>

where, p = $\frac{429}{1100}$ = 0.39

So, Null Hypothesis, $H_0$ : p = 39% {means that the proportion of adults who totally abstain from alcohol has not changed}

Alternate Hypothesis, $H_A$ : p $\neq$ 39% {means that the proportion of adults who totally abstain from alcohol has changed}

The test statistics that would be used here <u>One-sample z proportion statistics</u>;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of adults who totally abstain from alcohol = $\frac{352}{1100}$ = 0.32

n = sample of adults surveyed = 1100

So, <u><em>test statistics</em></u> = $\frac{0.32-0.39}{\sqrt{\frac{0.32(1-0.32)}{1100} } }$

= -4.976

The value of z test statistics is -4.976.

<em>Now, at 0.10 significance level the z table gives critical values of -1.645 and 1.645 for two-tailed test.</em><em> </em>

<em>Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the proportion of adults who totally abstain from alcohol has changed.

4 0

3 years ago