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krek1111 [17]
3 years ago
5

There are whole numbers that are not integers. True or false

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
7 0

Answer:

False

Step-by-step explanation:

All whole numbers are integers, along with all natural numbers being integers, but not all integers are whole numbers or natural numbers. ]

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Carly purchased 9.5 pints of ice cream for a party. If each guest will be served exactly 3/5 pints of ice cream, what is the gre
aivan3 [116]

Answer:

15.8333333333

Step-by-step explanation:

9.5 divied by 3.5

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3 years ago
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Quadratic and Ratios...​
lozanna [386]

Answer:

x=0

Step-by-step explanation:

in the quadratic formula to only have one solution

b^{2} -4ac=0 for ax^2+bx+c=0

so we have

ax^2+b=c

ax^2+(b-c)=0

if we insert in the formula we have

0^2-4(a)(b-c)=0\\\\-4ab+4ac=0\\\\4ac=4ab\\\\\frac{4ac}{4a} = \frac{4ab}{4a}\\\\c=b

ax^2+b=c\\\\ax^2=b-b\\\\ax^2=0 since a≠0

\frac{ax^2}{a}=\frac{0}{a}  \\\\x^2=0\\\\x=0

8 0
3 years ago
In October 1945, the Gallup organization asked 1,487 randomly sampled Americans, "Do you think we can develop a way to protect o
GalinKa [24]

Answer:

Yes. There is enough evidence to support the claim that the proportion of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945 is significantly greater than 0.5.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>In October 1945, the Gallup organization asked 1487 randomly sampled Americans, "Do you think we can develop a way to protect ourselves from atomic bombs in case other countries tried to use them against us?"</em><em> with 788 responding yes.</em><em> Did a majority of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945? Use the α=0.05 level of significance.</em>

This is a hypothesis test for a proportion.

The claim is that the proportion of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945 is significantly greater than 0.5.

Then, the null and alternative hypothesis are:

H_0: \pi=0.5\\\\H_a:\pi>0.5

The significance level is 0.05.

The sample has a size n=1487.

The sample proportion is p=0.53.

p=X/n=788/1487=0.53

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{1487}}\\\\\\ \sigma_p=\sqrt{0.000168}=0.013

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.53-0.5-0.5/1487}{0.013}=\dfrac{0.03}{0.013}=2.288

This test is a right-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z>2.288)=0.011

As the P-value (0.011) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of Americans feel the United States could develop a way to protect itself from atomic bombs in 1945 is significantly greater than 0.5.

8 0
3 years ago
1 / 4 + 1 / 3 + 7 / 10 =
Harlamova29_29 [7]
The answer should be 77/60
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3 years ago
1/4 + 1/2 divided by 3/8 I need the answer IMMEDIATELY!!! or im gonna get grounded ;-;
Bond [772]

.25+ .5= .75/ .375= 2

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3 years ago
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