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3 years ago

Hi! I could really use some help :)

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

8 0

Answer:

A=330

Step-by-step explanation:

im not sure which is the length width height

but if length is 10

width is 3/5

height is 15

A=2(wl+hl+hw)=2·(0.6·10+15·10+15·0.6)=330

then the area would be 330

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Random question what is 50+40-1*5plus 2

Scilla [17]

Answer:

87 i think

Step-by-step explanation:

I used a calculator

5 0

3 years ago

Read 2 more answers

Marianne wanted to know students' opinion on the new schedule at school. She survey's the first

Sonja [21]

I think it is a Convience Sampling.

6 0

3 years ago

Convert,the complex number into polar form: 4+4i

kow [346]

Z = a + bi
z = 4 + 4i

r² = a² + b²
r² = (4)² + (4)²
r² = 16 + 16
r² = 32
r = 4√(2)
r = 4(1.414)
r = 5.656

$cos\theta = \frac{a}{r}$
$cos\theta = \frac{4}{4\sqrt{2}}$
$cos\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$cos\theta = \frac{4\sqrt{2}}{4\sqrt{4}}$
$cos\theta = \frac{4\sqrt{2}}{4(2)}$
$cos\theta = \frac{4\sqrt{2}}{8}$
$cos\theta = \frac{\sqrt{2}}{2}$
$2(cos\theta) = 2(\frac{\sqrt{2}}{2})$
$2cos\theta = \sqrt{2}$
$2cos\theta = 1.414$

$sin\theta = \frac{b}{r}$
$sin\theta = \frac{4}{4\sqrt{2}}$
$sin\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$sin\theta = \frac{4\sqrt{2}}{4\sqrt{4}}$
$sin\theta = \frac{4\sqrt{2}}{4(2)}$
$sin\theta = \frac{4\sqrt{2}}{8}$
$sin\theta = \frac{\sqrt{2}}{2}$
$2(sin\theta) = 2(\frac{\sqrt{2}}{2})$
$2sin\theta = \sqrt{2}$
$2sin\theta = 1.414$

z = a + bi
z = rcosθ + (rsinθ)i
z = r(cosθ + i sinθ)

z = 4 + 4i
z = 5.656cosθ + (5.656sinθ)i
z = 5.656(cosθ + i sinθ)
z = 5.656(cos45 + i sin45)

$\theta = tan^{-1}\frac{b}{a}$
$\theta = tan^{-1}\frac{4}{4}$
$\theta = tan^{-1}(1)$
$\theta = 45$

The polar form of 4 + 4i is approximately equal to 5.656(cos45 + i sin45).

5 0

3 years ago

Leroi and Sylvia both put $200 in a savings account. Leroi decides he will put in an additional $40 each week. Sylvia decides to

Arlecino [84]

Answer:

First deposit - Same amount

Second deposit - Sylvia

Third deposit - Sylvia

Step-by-step explanation:

Leroi put $200 in their account. They will add $40 every week.

Deposit 1: $240

Deposit 2: $280

Deposit 3: $320

Sylvia put $200 in their account. They will add 20% of the amount in the account each week.

Deposit 1: $240 (20% of $200 is $40)

Deposit 2: $288 (20% of $240 is $48)

Deposit 3: $345.60 (20% of $288 is $57.60)

So, Sylvia and Leroi have the same amount of money in their accounts after the first deposit. After the second and third deposit, Sylvia has more money.

4 0

3 years ago

In a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer

stira [4]

Answer:

We conclude that 80% of patients stop smoking when given sustained care.

Step-by-step explanation:

We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.

Let p = percentage of patients stop smoking when given sustained care.

So, Null Hypothesis, $H_0$ : p = 80% {means that 80% of patients stop smoking when given sustained care}

Alternate Hypothesis, $H_A$ : p $\neq$ 80% {means that different from 80% of patients stop smoking when given sustained care}

The test statistics that would be used here One-sample z test for proportions;

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

where, $\hat p$ = sample proportion of patients who stop smoking when given sustained care = 80.2%

n = sample of patients = 192

So, the test statistics = $\frac{0.802-0.80}{\sqrt{\frac{0.802(1-0.802)}{192} } }$

= 0.07

The value of z test statistics is 0.07.

Also, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.07) = 1 - P(Z $\leq$ 0.07)

= 1 - 0.52790 = 0.4721

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 80% of patients stop smoking when given sustained care.

8 0

3 years ago