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OlgaM077 [116]
3 years ago
15

Winnie needs to spend $120 on two kinds of fancy cakes:

Mathematics
1 answer:
Ainat [17]3 years ago
3 0

Answer:

Winnie needs to buy 6 slices of Coffee cake.

Step-by-step explanation:

12 * 5= $60 (chocolate cake)

120 - 60= 60

since we know the coffee cake costs $10(per slice) we do:

60/ $10= 6 slices of coffee cake.

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-4+3x=8 solve plzzzz
jarptica [38.1K]

Answer:

x=4

Step-by-step explanation:

-4+3x=8

+4      +4

<u>3x</u>=<u>12</u>

3x  3x

x=4

3 0
3 years ago
Read 2 more answers
Please help,
Bess [88]

It is A bc (4)=7(1)-3

5 0
3 years ago
"Data from the World Health Organization in 2013 were used to predict the life expectancy of men in a country from the life expe
marshall27 [118]

Answer:

The predicted life expectancy of men in a country in which the life expectancy of women is 70 years is 65.33 years.

Step-by-step explanation:

The least square regression line is used to predict the value of the dependent variable from an independent variable.

The general form of a least square regression line is:

\hat y=\alpha +\beta x

Here,

<em>y</em> = dependent variable

<em>x</em> = independent variable

<em>α</em> = intercept

<em>β</em> = slope

The regression line to predict the life expectancy of men in a country from the life expectancy of women in that country is:

\hat y=3.73 +0.88 x

Compute the life expectancy of men in a country in which the life expectancy of women is 70 years as follows:

\hat y=3.73 +0.88 x

  =3.73 +(0.88\times 70)\\=3.73+61.6\\=65.33

Thus, the predicted life expectancy of men in a country in which the life expectancy of women is 70 years is 65.33 years.

6 0
3 years ago
An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so
alexgriva [62]

Solution :

Let p_1 and p_2  represents the proportions of the seeds which germinate among the seeds planted in the soil containing 3\% and 5\% mushroom compost by weight respectively.

To test the null hypothesis H_0: p_1=p_2 against the alternate hypothesis  H_1:p_1 \neq p_2 .

Let \hat p_1, \hat p_2 denotes the respective sample proportions and the n_1, n_2 represents the sample size respectively.

$\hat p_1 = \frac{74}{155} = 0.477419

n_1=155

$p_2=\frac{86}{155}=0.554839

n_2=155

The test statistic can be written as :

$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}

which under H_0  follows the standard normal distribution.

We reject H_0 at 0.05 level of significance, if the P-value or if |z_{obs}|>Z_{0.025}

Now, the value of the test statistics = -1.368928

The critical value = \pm 1.959964

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

                                     $=2 \times 0.085667$

                                     = 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$, so we fail to reject H_0 at 0.05 level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the \text{proportion} of the seeds that \text{germinate differs} with the percent of the \text{mushroom compost} in the soil.

8 0
3 years ago
Round 32,518 to the nearest hundreds
Afina-wow [57]
If I’m correct this is supposed to be a decimal ( here we use the dot not the comma ) but if so then 32.52 a good way to remember is 5 or more let it sore (meaning round up) four or less let it rest (meaning it stays the same)
6 0
3 years ago
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